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# Important Questions for CBSE Class 10 Maths Chapter 4 - Quadratic Equations

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 CBSE Class 10 Maths Important Questions Sl.No Chapter No Chapter Name 1 Chapter 1 Real Numbers 2 Chapter 2 Polynomials 3 Chapter 3 Pair of Linear Equations in Two Variables 4 Chapter 4 Quadratic Equations 5 Chapter 5 Arithmetic Progressions 6 Chapter 6 Triangles 7 Chapter 7 Coordinate Geometry 8 Chapter 8 Introduction to Trigonometry 9 Chapter 9 Some Applications of Trigonometry 10 Chapter 10 Circles 11 Chapter 11 Constructions 12 Chapter 12 Areas Related to Circles 13 Chapter 13 Surface Areas and Volumes 14 Chapter 14 Statistics 15 Chapter 15 Probability
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Important Questions for CBSE Class 10 Maths Chapter 4 - Quadratic Equations
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## Study Important Questions for Class 10 Mathematics Chapter 4 - Quadratic Equations

1. Solve by factorization

a. $4{{x}^{2}}-4{{a}^{2}}x+\left( {{a}^{4}}-{{b}^{4}} \right)=0$

Ans:

$4{{x}^{2}}-\left[ 2\left( {{a}^{2}}+{{b}^{2}} \right)+2\left( {{a}^{2}}-{{b}^{2}} \right) \right]\text{ }x+\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{a}^{2}}+{{b}^{2}} \right)=0$

$\Rightarrow 2x\left[ 2x-\left( {{a}^{2}}+{{b}^{2}} \right) \right]-\left( {{a}^{2}}-{{b}^{2}} \right)\left[ 2x-\left( {{a}^{2}}+{{b}^{2}} \right) \right]=0$

$\Rightarrow x=\dfrac{{{a}^{2}}+{{b}^{2}}}{2},x=\dfrac{{{a}^{2}}-{{b}^{2}}}{2}$

Therefore, $x=\dfrac{{{a}^{2}}+{{b}^{2}}}{2}\left( or \right)x=\dfrac{{{a}^{2}}-{{b}^{2}}}{2}$

b. ${{x}^{2}}+\left( \dfrac{a}{a+b}x+\dfrac{a+b}{a} \right)x+1=0$

Ans:

${{x}^{2}}+\left( \dfrac{a}{a+b}x+\dfrac{a+b}{a} \right)x+1=0$

$\Rightarrow {{x}^{2}}+\left( \dfrac{a}{a+b}x+\dfrac{a+b}{a}x+\dfrac{a}{a+b}.\dfrac{a+b}{a} \right)=0$

$\Rightarrow \left[ x+\dfrac{a}{a+b} \right]+\dfrac{a+b}{a}\left[ x+\dfrac{a}{a+b} \right]=0$

$\Rightarrow x=-\dfrac{a}{a+b},x=\dfrac{\left( -a+b \right)}{a},a+b=0$

c. $\dfrac{1}{a+b+x}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{x},\left( a+b\ne 0 \right)$

Ans:

$\dfrac{1}{a+b+x}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{x}$

$\Rightarrow \dfrac{1}{a+b+x}-\dfrac{1}{x}=\dfrac{1}{a}+\dfrac{1}{b}$

$\Rightarrow \dfrac{x-\left( a+b+x \right)}{x\left( a+b+x \right)}=\dfrac{a+b}{ab}$

$\Rightarrow \left( a+b \right)\left\{ x\left( a+b+x \right)+ab \right\}=0$

$\Rightarrow x\left( a+b+x \right)+ab=0$

$\Rightarrow {{x}^{2}}+ax+bx+ab=0$

$\Rightarrow \left( x+a \right)\left( x+b \right)=0$

$\Rightarrow x=-a\left( or \right)x=-b$

d. $\left( x-3 \right)\left( x-4 \right)\dfrac{34}{{{33}^{2}}}$

$\left( x-3 \right)\left( x-4 \right)=\dfrac{34}{{{33}^{2}}}$

$\Rightarrow {{x}^{2}}-7x+12=\dfrac{34}{{{33}^{2}}}$

$\Rightarrow {{x}^{2}}-7x+\dfrac{13034}{{{33}^{2}}}=0$

Ans:

$\Rightarrow {{x}^{2}}-7x+\dfrac{98}{33}\times \dfrac{133}{33}=0$

$\Rightarrow {{x}^{2}}-\dfrac{231}{33}x+\dfrac{98}{33}\times \dfrac{133}{33}=0$

$\Rightarrow {{x}^{2}}x-\dfrac{231}{33}x+\dfrac{98}{33}\times \dfrac{133}{33}=0$

$\Rightarrow {{x}^{2}}-\left( \dfrac{98}{33}+\dfrac{133}{33} \right)x+\dfrac{98}{33}\times \dfrac{133}{33}=0$

$\Rightarrow {{x}^{2}}-\dfrac{98}{33}x-\dfrac{133}{33}x+\dfrac{98}{33}\times \dfrac{133}{33}=0$

$\Rightarrow \left( x-\dfrac{98}{33} \right)x-\dfrac{133}{33}\left( x-\dfrac{98}{33} \right)=0$

$\Rightarrow \left( x-\dfrac{98}{33} \right)\left( x-\dfrac{133}{33} \right)=0$

$\Rightarrow x=\dfrac{98}{33}\left( or \right)x=\dfrac{133}{33}$

e. $\text{x=}\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x}}}x\text{ }\ne \text{ }2$

Ans:

$\text{x=}\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x}}}x\text{ }\ne \text{ }2\text{ }$

$\Rightarrow \text{ x=}\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x}}}$

$\Rightarrow \text{x=}\dfrac{1}{2-\dfrac{1}{2-\dfrac{\left( 2-x \right)}{4-2x-1}}}$

$\Rightarrow \text{x=}\dfrac{1}{2-\dfrac{2-x}{3-2x}}$

$\Rightarrow x=\dfrac{3-2x}{2\left( 3-2x \right)-\left( 2-x \right)}$

$\Rightarrow x=\dfrac{3-2x}{4-3x}$

$\Rightarrow 4x-3{{x}^{2}}\text{ }=3-2x$

$\Rightarrow 3{{x}^{2}}-6x+3=0\text{ }$

$\Rightarrow \text{ }{{\left( x-1 \right)}^{2}}\text{ }=0\text{ }$

$\Rightarrow x\text{ }=1,\text{ }1.$

2. By the method of completion of squares show that the equation $\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}+\mathbf{3x}+\mathbf{5}=\mathbf{0}$ has no real roots.

Ans:

$4{{x}^{2}}+3x+5=0$

$\Rightarrow {{x}^{2}}+\dfrac{3}{4}x+\dfrac{5}{4}=0$

$\Rightarrow {{x}^{2}}+\dfrac{3}{4}x+{{\left( \dfrac{3}{8} \right)}^{2}}=-\dfrac{5}{4}+\dfrac{9}{64}$

$\Rightarrow {{\left( x+\dfrac{3}{8} \right)}^{2}}=-\dfrac{71}{64}$

$\Rightarrow x+\dfrac{3}{8}=\sqrt{-\dfrac{71}{64}}$

Which is not a real number. Hence the equation has no real roots.

3. The sum of areas of two squares is 468m2. If the difference of their perimeters is 24cm, find the sides of the two squares.

Ans:

Let, the side of the larger square be x.

Let, the side of the smaller square be y.

${{x}^{2}}+{{y}^{2}}=468$

Cond. II 4x-4y = 24

$\Rightarrow xy=6$

$\Rightarrow x=6+y$

$\Rightarrow {{x}^{2}}+{{y}^{2}}=468$

$\Rightarrow {{\left( 6+y \right)}^{2}}+{{y}^{2}}=\text{ }468$

on solving we get

y = 12

⇒ x = (12+6) = 18 m

∴ The length of the sides of the two squares are 18m and 12m.

4. A dealer sells a toy for Rs.24 and gains as much percent as the cost price of the toy. Find the cost price of the toy.

Ans:

Let the C.P be x

∴Gain = x%

$\Rightarrow Gain\text{ }=x\dfrac{x}{100}$

S.P = C.P +Gain

SP = 24

$\Rightarrow x+\dfrac{{{x}^{2}}}{100}=24$

On solving we get x = 20 or x = -120 (reject this as cost cannot be negative)

∴ C.P of toy = Rs.20

5. A fox and an eagle lived at the top of a cliff of height 6m, whose base was at a distance of 10m from a point A on the ground. The fox descends the cliff and went straight to point A. The eagle flew vertically up to height x meters and then flew in a straight line to a point A, the distance traveled by each being the same. 3 Find the value of x.

Ans:

Distance traveled by the fox = distance traveled by the eagle

${(6+x)^2} + {(10)^2}$ = ${(16 – x)^2}$

on solving we get x = 2.72m.

6. A lotus is 2m above the water in a pond. Due to wind, the lotus slides on the side and only the stem completely submerges in the water at a distance of 10m from the original position. Find the depth of water in the pond.

Ans:

From,above figure,We can write as,

${(x+2)}^{2} = x^{2} + {10}^{2}$

$\Rightarrow x^{2} + 4x + 4$ = $x^{2} + 100$

$\Rightarrow 4x + 4 = 100$

$\Rightarrow x = 24$

Therefore, the depth of water in the pond is 24m.

7. Solve $x=\sqrt{6+\sqrt{6+\sqrt{6.....}}}$

Ans:

$x=\sqrt{6+\sqrt{6+\sqrt{6.....}}}$

$\Rightarrow x=\sqrt{6+x}$

$\Rightarrow {{x}^{2}}=6+x$

$\Rightarrow {{x}^{2}}-x-6=0$

$\Rightarrow \left( x\text{ }-3 \right)\left( x\text{ }+\text{ }2 \right)=0$

$\Rightarrow x\text{ }=\text{ }3$

As x cannot be negative x is not equal to 2.

8. The hypotenuse of a right triangle is 20m. If the difference between the length of the 4 other sides is 4m. Find the sides.

Ans:

From above figure,

${{x}^{2}}+{{y}^{2}}={{20}^{2}}$

${{x}^{2}}+{{y}^{2}}=400\text{ }$

$also\text{ }x-y=4$

$\Rightarrow x\text{ }=\text{ }404\text{ }+\text{ }y\text{ }$

$\Rightarrow {{\left( 4\text{ }+\text{ }y \right)}^{2}}+{{y}^{2}}=400$

$\Rightarrow 2{{y}^{2}}+8y-384=0$

$\Rightarrow \left( y\text{ }+\text{ }16 \right)\text{ }\left( y\text{ }\text{ }12 \right)=0$

$\Rightarrow y\text{ }=\text{ }12\text{ }\left( or \right)y=16\left( notpossible \right)$

∴sides are 12cm and 16cm

9. The positive value of k for which ${{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{Kx}\text{ }+\text{ }\mathbf{64}\text{ }=\text{ }\mathbf{0}\text{ }\And \text{ }{{\mathbf{x}}^{\mathbf{2}}}-~\mathbf{8x}\text{ }+\text{ }\mathbf{k}\text{ }=\text{ }\mathbf{0}$will have real roots.

Ans:

$\begin{array}{*{35}{l}} {{x}^{2}}+\text{ }Kx\text{ }+\text{ }64\text{ }=\text{ }0 \\ \Rightarrow {{b}^{2}}-4ac\text{ }\ge \text{ }0 \\ \Rightarrow {{K}^{2}}-256\text{ }\ge \text{ }0 \\ \Rightarrow K\text{ }\ge \text{ }16\text{ }or\text{ }K\text{ }\le \text{ }-\text{ }16\text{ }\ldots \ldots \ldots \ldots \ldots \text{ }\left( 1 \right) \\ {{x}^{2}}-8x\text{ }+\text{ }K\text{ }=\text{ }0 \\ 64\text{ }\text{ }4K\text{ }\ge \text{ }0 \\ \Rightarrow 4K\text{ }\le \text{ }64 \\ \Rightarrow K\text{ }\le \text{ }16\text{ }\ldots \ldots \ldots \ldots \ldots \text{ }\left( 2 \right) \\ From\text{ }\left( 1 \right)\text{ }\And \text{ }\left( 2 \right)\text{ }K\text{ }=\text{ }16 \\ \end{array}$

10. A teacher attempting to arrange the students for mass drill in the form of a solid square found that 24 students were left over. When he increased the size of the square by one student he found he was short of 25 students. Find the number of students.

Ans:

Let the side of the square be $x.$

No. of students = ${{x}^{2}}+\text{ }24$

New side = $x\text{ }+\text{ }1$

No. of students = ${{\left( x\text{ }+\text{ }1 \right)}^{2}}\text{ } - 25$

$\begin{array}{*{35}{l}} \Rightarrow {{x}^{2}}+\text{ }24=\text{ }{{\left( x\text{ }+\text{ }1 \right)}^{2}}\text{ }- 25 \\ \Rightarrow {{x}^{2}}+\text{ }24\text{ }=\text{ }{{x}^{2}}+\text{ }2x\text{ }+1\text{ }-25 \\ \Rightarrow 2x\text{ }=\text{ }48 \\ \Rightarrow x\text{ }=\text{ }24 \\ \end{array}$

∴ side of square = 24

No. of students = $576\text{ }+\text{ }24\text{ }=\text{ }600$

11. A pole has to be erected at a point on the boundary of a circular park of diameter 13m in such a way that the differences of its distances from two diametrically opposite fixed gates A $B on the boundary in 7m. Is it possible to do so? If the answer is yes at what distances from the two gates should the pole be erected? Ans: AB = 13 m BP = x$\begin{array}{*{35}{l}} \Rightarrow AP\text{ }-\text{ }BP\text{ }=\text{ }7  \\ \Rightarrow AP\text{ }=\text{ }x\text{ }+\text{ }7     \text{ }APQ  \\ \Rightarrow \left( 13 \right)^2\text{ }=\text{ }{{\left( x\text{ }+\text{ }7 \right)}^{2}}+\text{ }{{x}^{2}}  \\ \Rightarrow {{x}^{2}}+7x\text{ }\text{ }-60\text{ }=\text{ }0  \\ \Rightarrow \left( x\text{ }+\text{ }12 \right)\text{ }\left( x\text{ }\text{ }-5 \right)\text{ }=\text{ }0  \\ \Rightarrow x\text{ }=\text{ }-\text{ }12\text{ }\left( not\text{ }possible \right)\text{ }or\text{ }x\text{ }=\text{ }5  \\ \end{array}$∴Pole has to be erected at a distance of 5m from gate B & 12m from gate A. 12. If the roots of the equation$\left( \mathbf{a}-\mathbf{b} \right){{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{b}-\mathbf{c})\text{ }\mathbf{x}+\text{ }\left( \mathbf{c}\text{ }-\text{ }\mathbf{a} \right)=\text{ }\mathbf{0}$are equal. Prove that$\mathbf{2a}=\mathbf{b}\text{ }+\text{ }\mathbf{c}$. Ans:$\begin{array}{*{35}{l}} \left( ab \right){{x}^{2}}+\left( bc \right)\text{ }x+\left( ca \right)=0  \\ Given:2a=b+c  \\ {{B}^{2}}-4AC=0  \\ {{\left( bc \right)}^{2}}\left[ 4\left( ab \right)\left( c\text{ }\text{ }a \right) \right]=0  \\ \Rightarrow {{b}^{2}}-2bc+{{c}^{2}}\left[ 4\left( ac{{a}^{2}}bc+ab \right) \right]=0  \\ \Rightarrow {{b}^{2}}-2bc+{{c}^{2}}4ac+4{{a}^{2}}+4bc-\text{4}ab=0  \\ \Rightarrow {{b}^{2}}+2bc+{{c}^{2}}+4{{a}^{2}}4ac4ab=0  \\ \Rightarrow {{\left( \text{b}+c-2a \right)}^{2}}=0  \\ \Rightarrow b\text{ }+\text{ }c\text{ }=\text{ }2a  \\ \end{array}$Hence proved. 13. X and Y are centers of circles of radius 9cm and 2cm and XY = 17cm. Z is the center of a circle of radius 4 cm, which touches the above circles externally. Given that$\angle XYZ=90{}^\circ $, write an equation in r and solve it for r. Ans: Let r be the radius of the third circle XY = 17cm ⇒ XZ = 9 + r YZ = 2 + r$\begin{array}{*{35}{l}} {{\left( r\text{ }+\text{ }9 \right)}^{2}}+{{\left( r\text{ }+\text{ }2 \right)}^{2}}={{\left( 17 \right)}^{2}}  \\ \Rightarrow {{r}^{2}}+18r+81+{{r}^{2}}+4r+4=289  \\ \Rightarrow {{r}^{2}}+22r-204\text{ }=0  \\ \Rightarrow {{r}^{2}}+11r-102\text{ }=0 \\ \Rightarrow \left( r+17 \right)\left( r-6 \right)=0  \\ \Rightarrow r=-17\text{ }\left( not\text{ }possible \right)\text{ }or\text{ }r\text{ }=\text{ }6\text{ }cm  \\ \therefore radius\text{ }=\text{ }6cm.  \\ \end{array}$Level - 01 (01 Marks) 1. Check whether the following are quadratic equation or not i.$\left( x\text{ }-\text{ }3 \right)\text{ }\left( 2x\text{ }+\text{ }1 \right)\text{ }=\text{ }x\left( x\text{ }+\text{ }5 \right)$Ans: Yes, this is a quadratic equation as the highest power of x is 2. ii.$~{{\left( \mathbf{x}\text{ }+\text{ }\mathbf{2} \right)}^{\mathbf{2}}}=\text{ }\mathbf{2x}({{\mathbf{x}}^{\mathbf{2}}}-\text{ }\mathbf{1})$Ans: No, this is not a quadratic equation as the highest power of x is 1. 2. Solve by factorization method${{\mathbf{x}}^{\mathbf{2}}}-\mathbf{7x}+\mathbf{12}=\mathbf{0}$Ans:$x\text{ }=\text{ }3;\text{ }x\text{ }=\text{ }4$3. Find the discriminant${{\mathbf{x}}^{\mathbf{2}}}-\mathbf{3x}-\mathbf{10}=\text{ }\mathbf{0}$Ans:$D\text{ }=\text{ }49$(D = discriminant) 4. Find the nature of root$\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{3x}\text{ }-\text{ }\mathbf{4}\text{ }=\text{ }\mathbf{0}$Ans: root are real and unequal. 5. Find the value k so that quadratic equation$\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}-\mathbf{kx}+\mathbf{38}=\mathbf{0}$has equal root Ans:$5\text{ }\pm \text{ }18$6. Determine whether given value of x is a solution or not${{\mathbf{x}}^{\mathbf{2}}}-\mathbf{3x}-\mathbf{1}=\mathbf{0}:\mathbf{x}\text{ }=\text{ }\mathbf{1}$Ans: not a solution Level 2 (02 Marks) 7. Solve by quadratic equation$\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}-\text{ }\mathbf{24x}\text{ }-\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0}$by using quadratic formula. Ans:$1$8. Determine the value of for which the quadratic equation$\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}+\mathbf{3x}+\mathbf{k}=\text{ }\mathbf{0}$have both roots real. Ans:$k\le \dfrac{9}{8}x,x=\dfrac{3+\sqrt{10}}{4},\dfrac{3-\sqrt{10}}{4}$9. Find the roots of equation$\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}+\mathbf{x}-\mathbf{6}=\mathbf{0}$Ans:$x\text{ }=\text{ }2,x=\dfrac{3}{2}$10. Find the roots of equation$x-\dfrac{1}{x}=3;x\ne 0$Ans:$x=\dfrac{3}{2}$Level 3 (03 Marks) 1. The sum of the squares of two consecutive positive integers is 265. Find the integers. Ans: number are 11, 12 2. Divide 39 into two parts such that their product is 324. Ans: 27, 12 3. The sum of the number and its reciprocals is. Find the number. Ans:$4\dfrac{1}{4}$4. The length of a rectangle is 5cm more than its breadth if its area is 150 Sq. cm. Ans: 10cm, 15cm 5. The altitude of a right triangle is 7cm less than its base. If the hypotenuse is 13cm. Find the other two sides. Ans: 12cm and 5cm 1 Marks Questions 1. Which of the following is a quadratic equation?$ a){{x}^{3}}-2x-\sqrt{5}-x=0  b)3{{x}^{2}}-5x+9={{x}^{2}}-7x+3  c){{\left( x+\dfrac{1}{x} \right)}^{2}}=3\left( x+\dfrac{1}{x} \right)+4  d){{x}^{3}}+x+3=0 $Ans:$b)3{{x}^{2}}-5x+9={{x}^{2}}-7x+3$2. Factor of${{a}^{2}}{{x}^{2}}-3abx+2{{b}^{2}}=0$is$ a)\dfrac{2b}{a},\dfrac{b}{a}  b)\dfrac{3b}{a},\dfrac{a}{b}  c)\dfrac{b}{a},\dfrac{a}{b}  d)\dfrac{a}{b},\dfrac{a}{b} $Ans:$a)\dfrac{2b}{a},\dfrac{b}{a}$3. Which of the following have real roots?$ \text{a) 2}{{\text{x}}^{2}}+x-1=0  b)\text{ }{{\text{x}}^{2}}+x+1=0  c)\text{ }{{\text{x}}^{2}}-6x+6=0  d)\text{ 2}{{\text{x}}^{2}}+15x+30=0 $Ans:$c)\text{ }{{\text{x}}^{2}}-6x+6=0$4. Solve for x:$x=\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x}}} a)x=2  b)x=-1  c)x=1  d)x=3 $Ans:$b)x=-1$5. Solve by factorization$\sqrt{3}{{x}^{2}}+10x+7\sqrt{3}=0 a)x=-\sqrt{3},-\dfrac{7}{\sqrt{3}}  b)x=-\sqrt{3},\dfrac{7}{\sqrt{3}}  c)x=2,\dfrac{1}{2}  d)\pm 3 $Ans:$a)x=-\sqrt{3},-\dfrac{7}{\sqrt{3}}$6. The quadratic equation whose roots are 3 and -3 is$ a){{x}^{2}}-9=0  b){{x}^{2}}-3x-3=0  c){{x}^{2}}-2x+2=0  d){{x}^{2}}+9=0 $Ans:$a){{x}^{2}}-9=0$7. Discriminant of$-{{x}^{2}}+\dfrac{1}{2}x+\dfrac{1}{2}=0$is a)$-\dfrac{1}{2},1$b)$\dfrac{1}{2},1$c)$\dfrac{-1}{2},1$d)$\dfrac{1}{2},\dfrac{-1}{2}$Ans: (a)$-\dfrac{1}{2},1$8. For equal root,$kx\left( x-2 \right)+6=0$, value of k is a).$k=6$b).$k=3$c).$k=2$d.$k=8$Ans: (a)$k=6$9. Quadratic equation whose roots are$2+\sqrt{s},2-\sqrt{s}$is a).${{x}^{2}}-4x-1=0$b).${{x}^{2}}+4x+1=0$c).${{x}^{2}}+\left( x+\sqrt{5} \right)x-\left( 2\sqrt{5} \right)=0$d).${{x}^{2}}-4x+2=0$Ans: (a)${{x}^{2}}-4x-1=0$10. If$\alpha $and$\beta $are roots of the equation$3{{x}^{2}}+5x-7=0$then$\alpha \beta $equal to a).$\dfrac{7}{3}$b).$\dfrac{-7}{3}$c).$\dfrac{-5}{3}$d).$21$Ans: (b)$\dfrac{-7}{3}$2 Marks Questions 1. Solve the following problems given- i.$x{{\text{ }}^{2}}-45x+324=0$Ans:$x{{\text{ }}^{2}}-45x+324=0\Rightarrow {{x}^{2}}-36x-9x+324=0\Rightarrow x\left( x-36 \right)-9\left( x-36 \right)=0\Rightarrow \left( x-9 \right)\left( x-36 \right)=0\therefore x=36,9$ii.$x{{\text{ }}^{2}}-55x+750=0$Ans:$\Rightarrow x{{\text{ }}^{2}}-25x-30x+750=0\Rightarrow \text{ }x\left( x-25 \right)\text{ }\text{ }30\left( x-25 \right)=0\text{ }\Rightarrow \text{ }\left( x-30 \right)\left( x-25 \right)=0\therefore x=30,25$2. Find two numbers whose sum is$~27$and the product is$182$Ans: Let first number be x and let second number be$\left( 27-x \right)$According to given condition, the product of two numbers is$182$. Therefore,$x\left( 27-x \right)=182\Rightarrow \text{ }27x-x{{\text{ }}^{2}}=182\Rightarrow \text{ }x{{\text{ }}^{2}}-27x+182=0\Rightarrow \text{ }x{{\text{ }}^{2}}-27x+182=0\Rightarrow \text{ }x\left( x-14 \right)\text{ }\text{ }13\left( x-14 \right)=0\Rightarrow \text{ }\left( x-14 \right)\left( x-13 \right)=0\therefore \left( x-14 \right)\left( x-13 \right)=0$Therefore, the first number is equal to$14\text{ or 13}$And, second number is$=\text{ }27 \text{ }\text{ }x=27\text{ }\text{ }-14=13$or Second number$=\text{ }27\text{ }\text{ }-13=14$Therefore, two numbers are$13\text{ and }14$3. Find two consecutive positive integers, the sum of whose squares is$365$. Ans: Let first number be x and let second number be$\left( x+1 \right)$According to given condition${{x}^{2}}+{{\left( x+1 \right)}^{2}}=365\text{ }\left\{ {{\left( a+b \right)}^{2}}={{a}^{2}}+b{{\text{ }}^{2}}+2ab \right\}\Rightarrow \text{ }{{x}^{2}}+{{x}^{2}}+1+2x=365\Rightarrow \text{ }2{{x}^{2}}+2x\text{ }-364=0$Dividing equation by$2\Rightarrow \text{ }{{x}^{2}}+x\text{ }-182=0\Rightarrow \text{ }{{x}^{2}}+14x-13x\text{ }-182=0\Rightarrow \text{ }x\left( x+14 \right)\text{ }\text{ }-13\left( x+14 \right)=0\Rightarrow \text{ }\left( x+14 \right)\left( x-13 \right)=0\therefore x=13,-14$Therefore, first number$=13${We discard$-14$because it is negative number) Second number$=\text{ }x+1=13+1=14$Therefore, two consecutive positive integers are$13\,\text{and 14}$whose sum of squares is equal to$365$. 4. The altitude of a right triangle is$7$cm less than its base. If, hypotenuse is$13$cm. Find the other two sides. Ans: Let base of triangle be x cm and let altitude of triangle be (x−7) cm It is given that hypotenuse of triangle is 13 cm According to Pythagoras Theorem,$132\text{ }={{x}^{2}}+{{\left( x-7 \right)}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\Rightarrow \text{ }169={{x}^{2}}+{{x}^{2}}+49-14x\Rightarrow \text{ }169=2{{x}^{2}}-14x+49\text{ }\Rightarrow \text{ }2{{x}^{2}}-14x\text{ }120=0$Dividing equation by$2\Rightarrow \text{ }{{x}^{2}}-7x\text{ }60=0\Rightarrow \text{ }{{x}^{2}}-12x+5x\text{ }60=0\Rightarrow \text{ }x\left( x-12 \right)+5\left( x-12 \right)=0\Rightarrow \text{ }\left( x-12 \right)\left( x+5 \right)\therefore x=-5,12\text{ }$We discard$x=-5$because the length of the side of the triangle cannot be negative. Therefore, the base of triangle$=12$cm Altitude of triangle$=\left( x-7 \right)=12-7=5$cm 5. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was$3$more than twice the number of articles produced on that day. If, the total cost of production on that day was Rs.$90$, find the number of articles produced and the cost of each article. Ans: Let cost of production of each article be Rs$x$We are given total cost of production on that particular day$=\text{ }Rs\text{ }90$Therefore, total number of articles produced that day$=\text{ }\dfrac{90}{x}$According to the given conditions,$x=2\left( \dfrac{90}{x} \right)+3\Rightarrow x=\dfrac{180}{x}+3\Rightarrow x=\dfrac{180+3x}{x}\Rightarrow \text{ }{{x}^{2}}=180+3x\Rightarrow \text{ }{{x}^{2}}-3x\text{ }180=0\Rightarrow \text{ }{{x}^{2}}-15x+12x\text{ }180=0\Rightarrow \text{ }x\left( x-15 \right)+12\left( x-15 \right)=0~~~~\Rightarrow \text{ }\left( x-15 \right)\left( x+12 \right)=0~\therefore x=15,-12$Cost cannot be in negative; therefore, we discard$x=-12$Therefore$,\text{ }x=Rs15$which is the cost of production of each article. Number of articles produced on that particular day$=\dfrac{90}{15}=\text{ }6$6. In a class test, the sum of Shefali's marks in Mathematics and English is$30$. Had she got$2$marks more in Mathematics and$3$marks less in English, the product of their marks would have been$210$. Find her marks in the two subjects. Ans: Let Shefali's marks in Mathematics$=\text{ }x$Let Shefali's marks in English$=\text{ }30-x$If, she had got 2 marks more in Mathematics, her marks would be$=\text{ }x+2$If, she had got 3 marks less in English, her marks in English would be$=\text{ }30\text{ }\text{ }x-3\text{ }=\text{ }27-x$According to given condition:$\left( x+2 \right)\left( 27-x \right)=210\Rightarrow \text{ }27x-{{x}^{2}}+54-2x=210\Rightarrow \text{ }{{x}^{2}}-25x+156=0$Comparing quadratic equation${{x}^{2}}-25x+156=0$with general form$a{{x}^{2}}+bx+c=0$, We get$a=1,b=-25\text{ }and\text{ }c=156$Applying Quadratic Formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}x=\dfrac{25\pm \sqrt{{{\left( 25 \right)}^{2}}-4\left( 1 \right)\left( 156 \right)}}{2\times 1}\Rightarrow x=\dfrac{25\pm \sqrt{625-624}}{2}x=\dfrac{25\pm \sqrt{1}}{2}\Rightarrow x=\dfrac{25+1}{2},\dfrac{25-1}{2}\therefore x=13,12$Therefore, Shefali's marks in Mathematics$=\text{ }13\text{ }or\text{ }12$Shefali's marks in English$=\text{ }30\text{ }\text{ }x=30\text{ }\text{ }13=17$Or Shefali's marks in English$=\text{ }30\text{ }\text{ }x=30\text{ }\text{ }12=18$Therefore, her marks in Mathematics and English are$\left( 13,17 \right)\text{ }or\text{ }\left( 12,18 \right).$7. The diagonal of a rectangular field is$60$meters more than the shorter side. If, the longer side is$~30$meters more than the shorter side, find the sides of the field. Ans: Let shorter side of rectangle$=\text{ }x$meters Let diagonal of rectangle$=\text{ }\left( x+60 \right)$meters Let longer side of rectangle$=\text{ }\left( x+30 \right)$meters According to Pythagoras theorem,${{\left( x+60 \right)}^{2}}={{\left( x+30 \right)}^{2}}+{{x}^{2}}\Rightarrow \text{ }{{x}^{2}}+3600+120x={{x}^{2}}+900+60x+{{x}^{2}}~\Rightarrow \text{ }{{x}^{2}}-60x\text{ }2700=0{{x}^{2}}-60x\text{ }2700=0$Comparing equation${{x}^{2}}-60x\text{ }2700=0$with standard form$a{{x}^{2}}+bx+c=0$, We get$a=1,b=-60\text{ }and\text{ }c=-2700$Applying quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}x=\dfrac{60\pm \sqrt{{{\left( 60 \right)}^{2}}-4\left( 1 \right)\left( -2700 \right)}}{2\times 1}\Rightarrow x=\dfrac{60\pm \sqrt{3600+10800}}{2}\Rightarrow x=\dfrac{60\pm \sqrt{14400}}{2}=\dfrac{60\pm 120}{2}\Rightarrow x=\dfrac{60+120}{2},\dfrac{60-120}{2}\therefore x=90,-30$We ignore$-30$. Since length cannot be negative. Therefore,$x=90$which means length of shorter side$=90$meters And length of longer side$=\text{ }x+30\text{ }=\text{ }90+30=120$meters Therefore, length of sides is$90\text{ and }120$in meters. 8. The difference of squares of two numbers is$180$. The square of the smaller number is$8$times the larger number. Find the two numbers. Ans: Let smaller number$=\text{ }x$and let larger number$=\text{ }y$According to condition:${{y}^{2}}-{{x}^{2}}=180\text{ }\ldots \text{ }\left( 1 \right)$Also, we are given that square of smaller number is$8$times the larger number.$\Rightarrow \text{ }{{x}^{2}}=8y\text{ }\ldots \text{ }\left( 2 \right)$Putting equation (2) in (1), we get${{y}^{2}}-8y=180\Rightarrow \text{ }{{y}^{2}}-8y\text{ }180=0$Comparing equation${{y}^{2}}-8y\text{ }180=0$with general form$a{{y}^{2}}+by+c=0$, We get$a=1,b=-8\text{ and }c=-180$Using quadratic formula$y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}y=\dfrac{8\pm \sqrt{{{\left( -8 \right)}^{2}}-4\left( 1 \right)\left( -180 \right)}}{2\times 1}\Rightarrow y=\dfrac{8\pm \sqrt{64+720}}{2}\Rightarrow y=\dfrac{8\pm \sqrt{784}}{2}=\dfrac{8\pm 28}{2}\Rightarrow y=\dfrac{8+28}{2},\dfrac{8-28}{2}\therefore y=18,-10$Using equation (2) to find smaller number:$x{{\text{ }}^{2}}\text{ }=8y\Rightarrow \text{ }{{x}^{2}}=8y=8\times 18=144\Rightarrow \text{ }x=\pm 12$And,${{x}^{2}}=8y=8\times -10=-80${No real solution for$x$} Therefore, two numbers are$\left( 12,18 \right)\text{ }or\text{ }\left( -12,18 \right)$9. A train travels 360 km at a uniform speed. If, the speed had been 5 km/hr. more, it would have taken 1 hour less for the same journey. Find the speed of the train. Ans: Let the speed of the train = x km/hr If, speed had been 5km/hr more, train would have taken 1 hour less. So, according to this condition$\dfrac{360}{x}=\dfrac{360}{x+5}+1\Rightarrow 360\left( \dfrac{1}{x}-\dfrac{1}{x+5} \right)=1\Rightarrow 360\left( \dfrac{x+5-x}{x\left( x+5 \right)} \right)=1\Rightarrow \text{ }360\times 5={{x}^{2}}+5x\Rightarrow \text{ }{{x}^{2}}+5x\text{ }1800=0$Comparing equation${{x}^{2}}+5x\text{ }1800=0$with general equation$a{{x}^{2}}+bx+c=0$, We get$a=1,b=5\text{ and }c=-1800$Applying quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}x=\dfrac{-5\pm \sqrt{{{\left( 5 \right)}^{2}}-4\left( 1 \right)\left( -1800 \right)}}{2\times 1}\Rightarrow x=\dfrac{-5\pm \sqrt{25+7200}}{2}\Rightarrow x=\dfrac{-5\pm \sqrt{7225}}{2}=\dfrac{-5\pm 85}{2}\Rightarrow x=40,-45$Since the speed of train cannot be in negative. Therefore, we discard$x=-45$Therefore, speed of train$=\text{ }40$km/hr 10. Find the value of k for each of the following quadratic equations, so that they have two equal roots. i.$2{{x}^{2}}+kx+3=0$Ans:$2{{x}^{2}}+kx+3=0$We know that a quadratic equation has two equal roots only when the value of the discriminant is equal to zero. Comparing equation$2{{x}^{2}}+kx+3=0$with general quadratic equation$a{{x}^{2}}+bx+c=0$, we get$a=2,b=k\text{ }and\text{ }c=3$Discriminant$=\text{ }{{b}^{2}}-4ac={{k}^{2}}\text{ }4\left( 2 \right)\left( 3 \right)={{k}^{2}}-24$Putting discriminant equal to zero$~{{k}^{2}}\text{ }24=0\Rightarrow {{k}^{2}}=24\Rightarrow k=\pm \sqrt{24}=\pm 2\sqrt{6}\Rightarrow k=2\sqrt{6},-2\sqrt{6}$ii.$kx\left( x-2 \right)+6=0$Ans:$kx\left( x-2 \right)+6=0\Rightarrow \text{ }k{{x}^{2}}-2kx+6=0$Comparing quadratic equation$k{{x}^{2}}-2kx+6=0$with general form$a{{x}^{2}}+bx+c=0$, we get$a=k,b=\text{ }-2k\text{ }and\,c=6$Discriminant$=\text{ }{{b}^{2}}-4ac={{\left( -2k \right)}^{2}}\text{ }4\left( k \right)\left( 6 \right)=4{{k}^{2}}-24k$We know that two roots of quadratic equation are equal only if discriminant is equal to zero. Putting discriminant equal to zero$4{{k}^{2}}-24k=0\Rightarrow \text{ }4k\left( k-6 \right)=0\Rightarrow \text{ }k=0,6$The basic definition of quadratic equation says that quadratic equation is the equation of the form$a{{x}^{2}}+bx+c=0$, where$a\ne 0.$Therefore, in equation$k{{x}^{2}}-2kx+6=0$, we cannot have$k=0$. Therefore, we discard$k=0$. Hence the answer is$k=6$11. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is$800\text{ }{{m}^{2}}$. If so, find its length and breadth. Ans: Let breadth of rectangular mango grove$=\text{ }x$meters Let length of rectangular mango grove$=\text{ }2x$meters Area of rectangle = length × breadth$=\text{ }x\times \text{ }2x\text{ }=\text{ }2{{x}^{2}}{{m}^{2}}$According to given condition-$2{{x}^{2}}=800\Rightarrow \text{ }2{{x}^{2}}\text{ }800=0\Rightarrow \text{ }{{x}^{2}}\text{ }400=0$Comparing equation${{x}^{2}}\text{ }400=0$with general form of quadratic equation$a{{x}^{2}}+bx+c=0$, we get$a=1,b=0\text{ and }c=400$Discriminant$=\text{ }{{b}^{2}}-4ac={{\left( 0 \right)}^{2}}\text{ }4\left( 1 \right)\left( -400 \right)=1600$Discriminant is greater than 0 means that equation has two distinct real roots. Therefore, it is possible to design a rectangular grove. Applying quadratic formula,$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$to solve equation,$x=\dfrac{0\pm \sqrt{1600}}{2\times 1}=\dfrac{\pm 40}{2}=\pm 20\therefore x=20,-20$We discard negative value of$x$because breadth of rectangle cannot be in negative. Therefore,$x\text{ }=$breadth of rectangle$=\text{ }20$meters Length of rectangle$=\text{ }2x=2\times 20=40$meters 12. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is$~20$years. Four years ago, the product of their ages in years was$48$. Ans: Let age of first friend = x years and let age of second friend$=\text{ }\left( 20-x \right)$years Four years ago, age of first friend$=\text{ }\left( x-4 \right)$years Four years ago, age of second friend$=\text{ }\left( 20-x \right)-4\text{ }=\text{ }\left( 16-x \right)$years According to given condition,$\left( x-4 \right)\left( 16-x \right)=48~\Rightarrow \text{ }16x-{{x}^{2}}\text{ }64+4x=48\Rightarrow \text{ }20x-{{x}^{2}}\text{ }112=0\Rightarrow \text{ }{{x}^{2}}-20x+112=0$Comparing equation,${{x}^{2}}-20x+112=0$with general quadratic equation$a{{x}^{2}}+bx+c=0$, we get$a=1,b=-20\text{ }and\text{ }c=112$Discriminant$={{b}^{2}}-4ac={{\left( -20 \right)}^{2}}\text{ }4\left( 1 \right)\left( 112 \right)=400\text{ }\text{ }448=-48<0$The discriminant is less than zero which means we have no real roots for this equation. Therefore, the given situation is not possible. 13. Value of$x$for${{x}^{2}}-8x+15=0$is quadratic formula is a).$3,\text{ }2$b).$5,\text{ }2$c).$5,\text{ }3$d).$2,\text{ }3$Ans: (c)$5,\text{ }3$14. Discriminate of$\sqrt{3}{{x}^{2}}-2\sqrt{2}x-2\sqrt{3}=0$is a).$30$b).$31$c).$32$d).$35$Ans: (c)$32$15. Solve$12ab{{x}^{2}}-9{{a}^{2}}x+8{{b}^{2}}x-6ab=0$Ans:$12ab{{x}^{2}}-9{{a}^{2}}x+8{{b}^{2}}x-6ab=0\Rightarrow 3ax\left( 4bx-3x \right)+2b\left( 4bx-3x \right)=0\Rightarrow \left( 4bx-3x \right)\left( 3ax+2b \right)=0\Rightarrow 4bx-3a=0\,\,or\,\,3ax+2b=0\therefore x=\dfrac{3a}{4b}\,or\,x=-\dfrac{2b}{3a}$16. Solve for$x$by quadratic formula${{p}^{2}}{{x}^{2}}+\left( {{p}^{2}}-{{q}^{2}} \right)x-{{q}^{2}}=0$Ans:${{p}^{2}}{{x}^{2}}+\left( {{p}^{2}}-{{q}^{2}} \right)x-{{q}^{2}}=0a={{p}^{2}},\,b={{p}^{2}}-{{q}^{2}},\,c=-{{q}^{2}}D={{b}^{2}}-4ac=\left( {{p}^{2}}-{{q}^{2}} \right)-4\times {{p}^{2}}\left( -{{q}^{2}} \right)={{p}^{4}}+{{q}^{2}}-2{{p}^{2}}{{q}^{2}}+4{{p}^{2}}{{q}^{2}}={{\left( {{p}^{2}}+{{q}^{2}} \right)}^{2}}x=\dfrac{-b\pm \sqrt{D}}{29}=\dfrac{-\left( {{p}^{2}}{{q}^{2}} \right)\pm \sqrt{{{\left( {{p}^{2}}+{{q}^{2}} \right)}^{2}}}}{2\times {{p}^{2}}}=\dfrac{-{{p}^{2}}+{{q}^{2}}+{{p}^{2}}+{{q}^{2}}}{2{{p}^{2}}}or\,x=\dfrac{-{{p}^{2}}+{{q}^{2}}-{{p}^{2}}}{2{{p}^{2}}}x=\dfrac{2{{q}^{2}}}{2{{p}^{2}}}\,\,or\,x=\dfrac{-2{{q}^{2}}}{2{{p}^{2}}}x=\dfrac{{{q}^{2}}}{{{p}^{2}}}\,or\,x=-1$17. Find the value of k for which the quadratic equation$k{{x}^{2}}+2x+1=0$has real and distinct root Ans:$k{{x}^{2}}+2x+1=0a=k,\,b=2,\,c=1b={{b}^{2}}-4ac={{\left( 2 \right)}^{2}}-4\times k\times 1=4-4k$For real and distinct roots,$D > 04-4k > 0\Rightarrow -4k > -4\therefore k < 1 $18. If one root of the equations$2{{x}^{2}}+ax+3=0$is$1$, find the value of a. a).$=-4$b).$=-5$c).$=-3$d).$=-1$Ans: (b)$=-5$19. Find k for which the quadratic equation$4{{x}^{2}}-3kx+1=0$has equal root. 1.$=\pm \dfrac{3}{4}$2.$=\dfrac{3}{4}$3.$=\pm \dfrac{4}{3}$4.$=\dfrac{2}{3}$Ans: (c)$=\pm \dfrac{4}{3}$20. Determine the nature of the roots of the quadratic equation$9{{a}^{2}}{{b}^{2}}{{x}^{2}}-24abcdx+16{{c}^{2}}{{d}^{2}}=0$Ans:$D={{b}^{2}}-4ac={{\left( -24abcd \right)}^{2}}-4\times 9{{a}^{2}}{{b}^{2}}\times 16{{c}^{2}}{{d}^{2}}=576{{a}^{2}}{{b}^{2}}{{c}^{2}}-376{{a}^{2}}{{b}^{2}}{{c}^{2}}{{d}^{2}}=0$21. Find the discriminant of the equation$\left( x-1 \right)\left( 2x-1 \right)=0$Ans:$\left( x-1 \right)\left( 2x-1 \right)=0\Rightarrow 2{{x}^{2}}-x-2x+1=0\Rightarrow 2{{x}^{2}}-3x+1=0Here,a=2,\,b=-3,\,c=1D={{b}^{2}}-4ac={{\left( -3 \right)}^{2}}-4\times 2\times 1=9-8=1$22. Find the value of k so that$\left( x-1 \right)$is a factor of${{k}^{2}}{{x}^{2}}-2kx-3$. Ans: Let$P\left( x \right)={{k}^{2}}{{x}^{2}}-2kx-3P\left( 1 \right)={{k}^{2}}{{\left( 1 \right)}^{2}}-2k\left( 1 \right)-3\Rightarrow 0={{k}^{2}}-2k-3\Rightarrow {{k}^{2}}-3k+k-3\Rightarrow k\left( k-3 \right)+1\left( k-3 \right)=0\Rightarrow \left( k-3 \right)\left( k+1 \right)=0\therefore k=3\,or\,k=-1$23. The product of two consecutive positive integers is$306$. Represent these in quadratic equation. a).${{x}^{2}}+x-306=0$b).${{x}^{2}}-x-306=0$c).${{x}^{2}}+2x-106=0$d).${{x}^{2}}-x-106=0$Ans: (a)${{x}^{2}}+x-306=0$24. Which is a quadratic equation? a).${{x}^{2}}+x+2=0$b).${{x}^{3}}+{{x}^{2}}+2=0$c).${{x}^{4}}+{{x}^{2}}+2=0$d).$x+2=0$Ans: (a)${{x}^{2}}+x+2=0$25. The sum of two numbers is$16$. The sum of their reciprocals is$\dfrac{1}{3}$. Find the numbers. Ans: Let no. be$x$According to question,$\dfrac{1}{x}+\dfrac{1}{16-x}=\dfrac{1}{3}\Rightarrow \dfrac{16}{16x-{{x}^{2}}}=\dfrac{1}{3}\Rightarrow {{x}^{2}}-16x+48=0\Rightarrow {{x}^{2}}-12x-4x+48=0\therefore x=12\,or\,x=4$26. Solve for$x:\sqrt{217-x}=x-7$Ans:$\sqrt{217-x}=\left( x-7 \right)\Rightarrow 217-x={{x}^{2}}+49-14x\Rightarrow {{x}^{2}}-14x+x+49-217=0\Rightarrow {{x}^{2}}-13x-168=0\Rightarrow {{x}^{2}}-21x+8x-168=0\therefore x=21\,or\,x=-8$27. Solve for x by factorization:$x+\dfrac{1}{x}=11\dfrac{1}{11}$Ans:$\dfrac{{{x}^{2}}+1}{x}=\dfrac{122}{11}\Rightarrow 11{{x}^{2}}-12x+11=0\Rightarrow 11{{x}^{2}}-121x-1x+11=0\Rightarrow 11x\left( x-11 \right)-1\left( x-11 \right)=0\Rightarrow \left( 11x-1 \right)\left( x-11 \right)=0\therefore x=11\,or\,x=\dfrac{1}{11}$28. Find the ratio of the sum and product of the roots of$7{{x}^{2}}-12x+18=0$Ans:$7{{x}^{2}}-12x+18=0\alpha +\beta =\dfrac{-b}{a}=\dfrac{12}{7}\,and\alpha \beta =\dfrac{c}{a}=\dfrac{18}{17}\dfrac{\alpha +\beta }{\alpha \beta }=\dfrac{\dfrac{12}{7}}{\dfrac{18}{17}}=\dfrac{12}{7}\times \dfrac{17}{18}=\dfrac{34}{21}$29. If$\alpha $and$\beta $are the roots of the equation${{x}^{2}}+kx+12=0$, such that$\alpha -\beta =1$, then Ans:$\alpha +\beta =\dfrac{-k}{1},\alpha -\beta =1\alpha \beta =\dfrac{12}{1}{{\left( \alpha +\beta  \right)}^{2}}={{\left( \alpha -\beta  \right)}^{2}}+4\alpha \beta \Rightarrow {{\left( -k \right)}^{2}}={{\left( 1 \right)}^{2}}+4\times 12\Rightarrow {{k}^{2}}=49k=\pm 7$3 Marks Questions 1. Check whether the following are Quadratic Equations. i).${{\left( x+1 \right)}^{2}}=2\left( x-3 \right)$Ans:${{\left( x+1 \right)}^{2}}=2\left( x-3 \right)\,\left\{ {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\text{ } \right\}\Rightarrow \text{ }{{x}^{2}}+1+2x=2x\text{ }6\Rightarrow \text{ }{{x}^{2}}+7=0$Here, degree of equation is$2$. Therefore, it is a Quadratic Equation. ii).${{x}^{2}}-2x=\left( -2 \right)\left( 3-x \right)$Ans:${{x}^{2}}-2x=\left( -2 \right)\left( 3-x \right)\Rightarrow \text{ }{{x}^{2}}-2x=-6+2x\Rightarrow \text{ }{{x}^{2}}-2x-2x+6=0\Rightarrow \text{ }{{x}^{2}}-4x+6=0$Here, degree of equation is$2.$Therefore, it is a Quadratic Equation. iii).$\left( x-2 \right)\left( x+1 \right)=\left( x-1 \right)\left( x+3 \right)$Ans: (x−2)(x+1)=(x−1)(x+3)$\Rightarrow \text{ }{{x}^{2}}+x-2x\text{ }2={{x}^{2}}+3x\text{ }x\text{ }3=0\Rightarrow \text{ }{{x}^{2}}+x-2x\text{ }2-{{x}^{2}}-3x+x+3=0\Rightarrow \text{ }x-2x\text{ }2-3x+x+3=0\Rightarrow \text{ }-3x+1=0$Here, degree of equation is$1.$Therefore, it is not a Quadratic Equation. iv).$~~\left( x-3 \right)\left( 2x+1 \right)=x\left( x+5 \right)$Ans:$\left( x-3 \right)\left( 2x+1 \right)=x\left( x+5 \right)\Rightarrow \text{ }2{{x}^{2}}+x-6x\text{ }3={{x}^{2}}+5x\Rightarrow \text{ }2{{x}^{2}}+x-6x\text{ }3-{{x}^{2}}-5x=0\Rightarrow \text{ }{{x}^{2}}-10x\text{ }3=0$Here, degree of equation is 2. Therefore, it is a quadratic equation. v).$\left( 2x-1 \right)\left( x-3 \right)=\left( x+5 \right)\left( x-1 \right)$Ans:$\left( 2x-1 \right)\left( x-3 \right)=\left( x+5 \right)\left( x-1 \right)\text{ }\Rightarrow \text{ }2{{x}^{2}}-6x\text{ }x+3={{x}^{2}}\text{ }x+5x\text{ }5\Rightarrow \text{ }{{x}^{2}}-11x+8=0$Here, degree of Equation is$2$. Therefore, it is a Quadratic Equation. vi).${{x}^{2}}+3x+1={{\left( x-2 \right)}^{2}}$Ans:${{x}^{2}}+3x+1={{\left( x-2 \right)}^{2}}\,\,\,\,\,\text{ }\left\{ {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} \right\}\Rightarrow \text{ }{{x}^{2}}\text{ }+3x+1={{x}^{2}}+4-4x\Rightarrow \text{ }{{x}^{2}}+3x+1-{{x}^{2}}+4x\text{ }4=0\Rightarrow \text{ }7x\text{ }3=0$Here, degree of equation is$~1$. Therefore, it is not a Quadratic Equation. vii).${{\left( x+2 \right)}^{3}}=2x\left( {{x}^{2}}-1 \right)$Ans:${{\left( x+2 \right)}^{3}}=2x\left( {{x}^{2}}-1 \right)\text{ }\,\,\,\,\,\,\,\left\{ {{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right) \right\}\Rightarrow \text{ }{{x}^{3}}\text{ }+{{2}^{3}}\text{ }+3\left( x \right)\left( 2 \right)\left( x+2 \right)=2x\left( {{x}^{2}}-1 \right)\Rightarrow \text{ }{{x}^{3}}+8+6x\left( x+2 \right)=2{{x}^{3}}-2x\Rightarrow \text{ }2{{x}^{3}}-2x-{{x}^{3}}\text{ }8-6{{x}^{2}}-12x=0\Rightarrow \text{ }{{x}^{3}}-6{{x}^{2}}-14x\text{ }8=0$Here, degree of Equation is 3. Therefore, it is not a quadratic Equation. viii).${{x}^{3}}-4{{x}^{2}}\text{ }x+1={{\left( x-2 \right)}^{3}}\text{ }$Ans:${{x}^{3}}-4{{x}^{2}}\text{ }x+1={{\left( x-2 \right)}^{3}}\,\,\,\,\text{ }\left\{ {{\left( a-b \right)}^{3}}\text{ }={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right) \right\}\Rightarrow \text{ }{{x}^{3}}-4{{x}^{2}}\text{ }x+1={{x}^{3}}-{{2}^{3}}\text{ }3\left( x \right)\left( 2 \right)\left( x-2 \right)\Rightarrow \text{ }-4{{x}^{2}}\text{ }x+1=-8-6{{x}^{2}}+12x\Rightarrow \text{ }2{{x}^{2}}-13x+9=0$Here, degree of Equation is$2$. Therefore, it is a Quadratic Equation. 2. Represent the following situations in the form of Quadratic Equations: i). The area of the rectangular plot is 528$m^2$. The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot. Ans: We are given that area of a rectangular plot is$528{{m}^{2}}$Let the breadth of the rectangular plot be$~x$meters Length is one more than twice its breadth Therefore, length of rectangular plot is$\left( 2x+1 \right)$meters Area of rectangle$=$length$\times $breadth$\Rightarrow \text{ }528=x\left( 2x+1 \right)\Rightarrow \text{ }528=2{{x}^{2}}+x\Rightarrow \text{ }2{{x}^{2}}+x\text{ }528=0$This is a Quadratic Equation. ii). The product of two consecutive numbers is 306. We need to find the integers. Ans: Let two consecutive numbers be$x\,\text{and}\,\left( x+1 \right).$It is given that x(x+1) = 306$\Rightarrow \text{ }{{x}^{2}}+x=306\Rightarrow \text{ }{{x}^{2}}+x\text{ }306=0$This is a Quadratic Equation. iii). Rohan's mother is 26 years older than him. The product of their ages (in years) after 3 years will be 360. We would like to find Rohan's present age. Ans: Let present age of Rohan$=\text{ }x$years Let present age of Rohan's mother$=\text{ }\left( x\text{ }+26 \right)$years Age of Rohan after$~3$years$=\text{ }\left( x+3 \right)$years Age of Rohan's mother after$~3$years$=\text{ }x+26+3\text{ }=\text{ }\left( x+29 \right)$years According to given condition:$\left( x+3 \right)\left( x+29 \right)=360\Rightarrow \text{ }{{x}^{2}}+29x+3x+87=360\Rightarrow \text{ }{{x}^{2}}+32x\text{ }273=0$This is a Quadratic Equation. iv). A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train. Ans: Let the speed of the train be$x$km/h Time taken by train to cover 480 km$=\text{ }480x$hours If, speed had been 8km/h less than time taken would be$\left( 480x-8 \right)$hours. According to given condition, if speed had been$8$km/h less than time taken is$3$hours less. Therefore,$480x\text{ }8=480x+3\Rightarrow \text{ }480\left( 1x\text{ }8-1x \right)=3\Rightarrow \text{ }480\left( x\text{ }x+8 \right)\text{ }\left( x \right)\text{ }\left( x-8 \right)=3\Rightarrow \text{ }480\times 8=3\left( x \right)\left( x-8 \right)~\Rightarrow \text{ }3840=3{{x}^{2}}-24x$Dividing equation by$3$, we get$\Rightarrow \text{ }{{x}^{2}}-8x\text{ }1280=0$This is a Quadratic Equation. 3. Find the roots of the following Quadratic Equations by factorization. i).${{x}^{2}}-3x\text{ }10=0$Ans:${{x}^{2}}-3x\text{ }10=0\Rightarrow \text{ }{{x}^{2}}-5x+2x\text{ }10=0\Rightarrow \text{ }x\left( x-5 \right)+2\left( x-5 \right)=0\Rightarrow \text{ }\left( x-5 \right)\left( x+2 \right)=0\Rightarrow \text{ }x=5,-2$ii).$2{{x}^{2}}+x\text{ }6=0$Ans:$2{{x}^{2}}+x\text{ }6=0\Rightarrow \text{ }2{{x}^{2}}+4x-3x\text{ }6=0\Rightarrow \text{ }2x\left( x+2 \right)\text{ }\text{ }3\left( x+2 \right)=0\Rightarrow \text{ }\left( 2x-3 \right)\left( x+2 \right)=0\Rightarrow x=\dfrac{3}{2},2$iii).$\sqrt{2}{{x}^{2}}+7x+5\sqrt{2}=0$Ans:$\sqrt{2}{{x}^{2}}+7x+5\sqrt{2}=0\Rightarrow \sqrt{2}{{x}^{2}}+2x+5x+5\sqrt{2}=0\Rightarrow \sqrt{2}{{x}^{2}}\left( x+\sqrt{2} \right)+5\left( x+\sqrt{2} \right)=0\Rightarrow \left( \sqrt{2}x+5 \right)\left( x+\sqrt{2} \right)=0\Rightarrow x=\dfrac{-5}{\sqrt{2}},-\sqrt{2}\Rightarrow x=\dfrac{-5}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}},-\sqrt{2}\Rightarrow x=\dfrac{-5\sqrt{2}}{2},-\sqrt{2}$iv).$2{{x}^{2}}-x+\dfrac{1}{8}=0$Ans:$2{{x}^{2}}-x+\dfrac{1}{8}=0\Rightarrow \dfrac{16{{x}^{2}}-8x+1}{8}=0\Rightarrow \text{ }16{{x}^{2}}-8x+1=0\Rightarrow \text{ }16{{x}^{2}}-4x-4x+1=0\Rightarrow \text{ }4x\left( 4x-1 \right)\text{ }\text{ }1\left( 4x-1 \right)=0\Rightarrow \text{ }\left( 4x-1 \right)\left( 4x-1 \right)=0\Rightarrow \text{ }x=\text{ }\dfrac{1}{4},\dfrac{1}{4}$v).$100{{x}^{2}}-20x+1=0$Ans:$100{{x}^{2}}-20x+1=0\Rightarrow \text{ }100{{x}^{2}}-10x-10x+1=0\Rightarrow \text{ }10x\left( 10x-1 \right)\text{ }\text{ }1\left( 10x-1 \right)=0\Rightarrow \text{ }\left( 10x-1 \right)\left( 10x-1 \right)=0\therefore \,x=\dfrac{1}{10},\dfrac{1}{10}$4. Find the roots of the following equations: i).$\dfrac{x-1}{x}=3,x\ne 0$Ans:$x-\dfrac{1}{x}=3\,\,where\,x\ne 0\Rightarrow \dfrac{{{x}^{2}}-1}{x}=3\Rightarrow \text{ }x{{\text{ }}^{2}}\text{ }1=3x\Rightarrow \text{ }{{x}^{2}}-3x\text{ }1=0$Comparing equation${{x}^{2}}-3x\text{ }1=0$with general form$a{{x}^{2}}+bx+c=0$, We get$a=1,b=-3\text{ }and\text{ }c=-1$Using quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$to solve equation,$x=\dfrac{3\pm \sqrt{{{\left( 3 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)}}{2\times 1}\Rightarrow x=\dfrac{3\pm \sqrt{13}}{2}\Rightarrow x=\dfrac{3+\sqrt{13}}{2},\dfrac{3-\sqrt{13}}{2}$(ii).$\dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30},x\ne -4,7$Ans:$\dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30}\,where\,x\ne -4,7\Rightarrow \text{ }-30={{x}^{2}}-7x+4x28\Rightarrow \text{ }{{x}^{2}}-3x+2=0$Comparing equation$~{{x}^{2}}-3x+2=0$with general form$a{{x}^{2}}+bx+c=0$, We get$a=1,b=-3\text{ }and\text{ }c=2$Using quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$to solve equation,$x=\dfrac{3\pm \sqrt{{{\left( 3 \right)}^{2}}-4\left( 1 \right)\left( 2 \right)}}{2\times 1}\Rightarrow x=\dfrac{3\pm \sqrt{1}}{2}\Rightarrow x=\dfrac{3+\sqrt{1}}{2},\dfrac{3-\sqrt{1}}{2}\Rightarrow x=2,1$5. The sum of reciprocals of Rehman's ages (in years)$3$years ago and$5$years from now is$13$. Find his present age. Ans: Let present age of Rehman$=\text{ }x$years Age of Rehman$3$years ago$=\text{ }\left( x-3 \right)$years. Age of Rehman after$5$years$=\text{ }\left( x+5 \right)$years According to the given condition:$\dfrac{1}{x-3}+\dfrac{1}{x+5}=\dfrac{1}{3}\Rightarrow \dfrac{\left( x+5 \right)+\left( x-3 \right)}{\left( x-3 \right)\left( x+5 \right)}=\dfrac{1}{3}\Rightarrow \text{ }3\left( 2x+2 \right)\text{ }=\left( x-3 \right)\left( x+5 \right)\text{ }\Rightarrow \text{ }6x+6={{x}^{2}}-3x+5x-15\Rightarrow \text{ }{{x}^{2}}-4x\text{ }15\text{ }\text{ }6=0\Rightarrow {{x}^{2}}-4x\text{ }21=0$Comparing quadratic equation x${{x}^{2}}-4x\text{ }21=0$with general form$a{{x}^{2}}+bx+c=0$, We get$a=1,b=-4\text{ }and\text{ }c=-21$Using quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}x=\dfrac{4\pm \sqrt{{{\left( 4 \right)}^{2}}-4\left( 1 \right)\left( -21 \right)}}{2\times 1}\Rightarrow x=\dfrac{4\pm \sqrt{16+84}}{2}\Rightarrow x=\dfrac{4\pm \sqrt{100}}{2}=\dfrac{4\pm 10}{2}\Rightarrow x=\dfrac{4+10}{2},\dfrac{4-10}{2}\therefore x=7,-3$We discard$x=-3$.Since age cannot be in negative. Therefore, present age of Rehman is$7$years 6. Two water taps together can fill a tank in$9\dfrac{3}{8}$hours. The tap of larger diameter takes$10$hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. Ans: Let time taken by tap of smaller diameter to fill the tank$=\text{ }x$hours Let time taken by tap of larger diameter to fill the tank$=\text{ }\left( x\text{ }10 \right)$hours It means that tap of smaller diameter fills${{\dfrac{1}{x}}^{th}}$part of tank in$1$hour. … (1) And, tap of larger diameter fills${{\dfrac{1}{x-10}}^{th}}$part of tank in$1$hour. … (2) When two taps are used together, they fill tank in$~758$hours In 1 hour, they fill${{\dfrac{8}{75}}^{th}}$part of tank$\left[ \dfrac{1}{\dfrac{75}{8}}=\dfrac{8}{75} \right]$… (3) From (1), (2) and (3),$\dfrac{1}{x}+\dfrac{1}{x-10}=\dfrac{8}{75}\Rightarrow \dfrac{x-10+x}{x\left( x-10 \right)}=\dfrac{8}{75}\Rightarrow \text{ }75\left( 2x-10 \right)=8\left( {{x}^{2}}-10x \right)\Rightarrow \text{ }150x\text{ }750=8{{x}^{2}}-80x\Rightarrow \text{ }8{{x}^{2}}\text{ }-80x-150x+750=0$Comparing equation$4{{x}^{2}}\text{ }-115x+375=0$with general equation$a{{x}^{2}}+bx+c=0$, We get$a=4,b=-115\,and\,c=375$Applying quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}x=\dfrac{115\pm \sqrt{{{\left( -115 \right)}^{2}}-4\left( 4 \right)\left( 375 \right)}}{2\times 4}\Rightarrow x=\dfrac{115\pm \sqrt{13225-6000}}{8}\Rightarrow \dfrac{115\pm \sqrt{7225}}{8}\Rightarrow \dfrac{115+85}{8},\dfrac{115-85}{8}\therefore x=25,3.75$Time taken by larger tap$=\text{ }x\text{ }10=3.75\text{ }\text{ }10=-6.25$hours Time cannot be in negative. Therefore, we ignore this value. Time taken by larger tap$=\text{ }x\text{ }10=25\text{ }\text{ }10=15$hours Therefore, time taken by larger tap is$15$hours and time taken by smaller tap is$25$hours. 7. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them. i).$2{{x}^{2}}\text{ }3x\text{ }+\text{ }5\text{ }=\text{ }0$Ans:$2{{x}^{2}}\text{ }3x\text{ }+\text{ }5\text{ }=\text{ }0$Comparing this equation with general equation$a{{x}^{2}}+bx+c=0$, We get$a=2,b=-3\text{ }and\text{ }c=5$Discriminant$=\text{ }{{b}^{2}}-4ac={{\left( -3 \right)}^{2}}\text{ }4\left( 2 \right)\left( 5 \right)=9\text{ }\text{ }40=-31$Discriminant is less than 0 which means the equation has no real roots. ii).$3{{x}^{2}}-4\sqrt{3}x+4=0$Ans:$3{{x}^{2}}-4\sqrt{3}x+4=0$Comparing this equation with general equation$a{{x}^{2}}+bx+c=0$, We get$a=3,b=-4\sqrt{3}\text{ }and\text{ }c=4$Discriminant$=\text{ }{{b}^{2}}-4ac={{\left( -4\sqrt{3} \right)}^{2}}-4\left( 3 \right)\left( 4 \right)=48\text{ }\text{ }48=0$Discriminant is equal to zero which means equations have equal real roots. Applying quadratic$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$to find roots,$x=\dfrac{4\sqrt{3}\pm \sqrt{0}}{6}=\dfrac{2\sqrt{3}}{3}$Because, equation has two equal roots, it means$x=\dfrac{2\sqrt{3}}{3},\dfrac{2\sqrt{3}}{3}$iii).$2{{x}^{2}}\text{ }+6x\text{ }+\text{ }3\text{ }=\text{ }0$Ans:$2{{x}^{2}}\text{ }+6x\text{ }+\text{ }3\text{ }=\text{ }0$Comparing equation with general equation$a{{x}^{2}}+bx+c=0$, We get$a=2,b=-6,\text{ }and\text{ }c=3$Discriminant$=\text{ }{{b}^{2}}-4ac={{\left( -6 \right)}^{2}} - \text{ }4\left( 2 \right)\left( 3 \right)=36\text{ }-\text{ }24=12$Value of the discriminant is greater than zero. Therefore, the equation has distinct and real roots. Applying quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$to find roots,$x=\dfrac{6\pm \sqrt{12}}{4}=\dfrac{6\pm 2\sqrt{3}}{4}\Rightarrow x=\dfrac{3\pm \sqrt{3}}{2}\Rightarrow x=\dfrac{3+\sqrt{3}}{2},\dfrac{3-\sqrt{3}}{2}$8. If$-4$is a root of the quadratic equation and the quadratic equation${{x}^{2}}+px-4$has equal root, find the value of$k.$Ans:$-4$is root of${{x}^{2}}+px-4=0\therefore {{\left( -4 \right)}^{2}}+p\left( -4 \right)-4=0\Rightarrow 16-4p-4=0\Rightarrow -4p=-12\Rightarrow p=3{{x}^{2}}+px+k=0$(Given)${{x}^{2}}+3x+k=0D={{b}^{2}}-4ac\Rightarrow 0={{\left( 3 \right)}^{2}}-4\times 1\times k$[For equal roots D = 0]$\Rightarrow 4k=9\Rightarrow k=\dfrac{9}{4}$9. Solve for$x:{{5}^{x+1}}+{{5}^{1-x}}=26$Ans:${{5}^{x+1}}+{{5}^{1-x}}=26{{5}^{x}}{{.5}^{1}}+{{5}^{1}}{{.5}^{-x}}=26\Rightarrow {{5}^{x}}.5+\dfrac{{{5}^{1}}}{{{5}^{x}}}=26$Put${{5}^{x}}=y\dfrac{5y}{1}+\dfrac{5}{y}=26\Rightarrow 5{{y}^{2}}-26y+5=0\Rightarrow 5{{y}^{2}}-25y-y+5=0\Rightarrow 5y\left( y-5 \right)-1\left( y-5 \right)=0\Rightarrow \left( y-5 \right)\left( 5y-1 \right)=0\Rightarrow y=5$or$y=\dfrac{1}{5}$But${{5}^{x}}={{5}^{1}}$and${{5}^{x}}=\dfrac{1}{5}\Rightarrow x=1$and${{5}^{x}}={{5}^{-1}}\Rightarrow x=-1$10.$\dfrac{1}{p+q+x}=\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{x}$solve for$x$by factorization method. Ans:$\dfrac{1}{p+q+x}=\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{x}\Rightarrow \dfrac{1}{p+q+x}-\dfrac{1}{x}=\dfrac{1}{p}+\dfrac{1}{q}\Rightarrow \dfrac{x-p-q-x}{{{x}^{2}}+px+qx}=\dfrac{p+q}{pq}\Rightarrow \dfrac{-\left( p+q \right)}{{{x}^{2}}+px+qx}=\dfrac{p+q}{pq}\Rightarrow \dfrac{-1}{{{x}^{2}}+px+qx}=\dfrac{1}{pq}\Rightarrow {{x}^{2}}+px+qx=-pq\Rightarrow {{x}^{2}}+px+qx+pq=0\Rightarrow x\left( x+p \right)+q\left( x+p \right)=0\Rightarrow \left( x+p \right)\left( x+q \right)=0\therefore x=-p\,or\,x=-q$11.$5{{x}^{2}}-6x-2=0$, solve for$x$by the method of completing the square. Ans:$5{{x}^{2}}-6x-2=0\Rightarrow {{x}^{2}}-\dfrac{6}{5}x-\dfrac{2}{5}=0\Rightarrow {{x}^{2}}-\dfrac{6}{5}x+{{\left( \dfrac{3}{5} \right)}^{2}}-\dfrac{2}{5}=0\Rightarrow {{\left( x-\dfrac{3}{5} \right)}^{2}}=\dfrac{9}{25}+\dfrac{2}{5}\Rightarrow {{\left( x-\dfrac{3}{5} \right)}^{2}}=\dfrac{9+10}{25}\Rightarrow {{\left( x-\dfrac{3}{5} \right)}^{2}}=\dfrac{19}{25}\Rightarrow x-\dfrac{3}{5}=\pm \dfrac{\sqrt{19}}{5}\Rightarrow x=\dfrac{3}{5}\pm \dfrac{\sqrt{19}}{5}\Rightarrow x=\dfrac{3+\sqrt{19}}{5}\,or\,x=\dfrac{3-\sqrt{19}}{5}$12. Solve for$x:{{a}^{2}}{{b}^{2}}{{x}^{2}}+{{b}^{2}}x-{{a}^{2}}x-1=0$Ans:${{a}^{2}}{{b}^{2}}{{x}^{2}}+{{b}^{2}}x-{{a}^{2}}x-1=0\Rightarrow {{b}^{2}}x\left( {{a}^{2}}x+1 \right)-1\left( {{a}^{2}}x-1 \right)=0\Rightarrow \left( {{a}^{2}}x+1 \right)\left( {{b}^{2}}x-1 \right)=0\therefore x=\dfrac{-1}{{{a}^{2}}}\,or\,x=\dfrac{-1}{{{b}^{2}}}$13. Using quadratic formula, solve for$x:9{{x}^{2}}-9\left( a+b \right)x+\left( 2{{a}^{2}}+5ab+2{{b}^{2}} \right)=0$Ans:$D={{b}^{2}}-4ac={{\left( -9\left( a+b \right) \right)}^{2}}-4\times 9\times \left( 2{{a}^{2}}+5ab+2a{{b}^{2}} \right)=81{{\left( a+b \right)}^{2}}-36\left( 2{{a}^{2}}+5ab+2{{b}^{2}} \right)=9\left[ 9\left( {{a}^{2}}+{{b}^{2}}+2ab-8{{a}^{2}}-20ab-8{{b}^{2}} \right) \right]=9\left[ {{a}^{2}}+{{b}^{2}}-2ab \right]=9{{\left( a-b \right)}^{2}}x=\dfrac{-b\pm \sqrt{D}}{2a}=\dfrac{9\left( a+b \right)\pm \sqrt{9{{\left( a-b \right)}^{2}}}}{2\times 9}\Rightarrow x=3\dfrac{\left[ 3\left( a+b \right)\pm \left( a-b \right) \right]}{2\times 9}\Rightarrow x=\dfrac{\left( 3a+3b \right)\pm \left( a-b \right)}{6}\Rightarrow x=\dfrac{3a+3b+a-b}{6}\,or\,x=\dfrac{3a+3b+a-b}{6}\Rightarrow x=\dfrac{4a+2b}{6}\,or\,x=\dfrac{4a+2b}{6}\Rightarrow x=\dfrac{2a+b}{3}\,or\,x=\dfrac{2a+b}{3}$14. In a cricket match, Kapil took one wicket less than twice the number of wickets taken by Ravi. If the product of the numbers of wickets taken by these two is 15, find the number of wickets taken by each. Ans: Let no. of wicket taken by Ravi$=x$of wicket taken by Kapil$=2x-1$According to question,$\left( 2x-1 \right)x=15\Rightarrow 2{{x}^{2}}-x-15=0\therefore x=3\,or\,x=\dfrac{-5}{2}$(Neglects) So, no. of wickets taken by Ravi is$x=3$15. The sum of a number and its reciprocal is$\dfrac{17}{4}$. Find the number. Ans: Let no. be$x$According to question,$\dfrac{x}{1}+\dfrac{1}{x}=\dfrac{17}{4}\Rightarrow \dfrac{{{x}^{2}}+1}{x}=\dfrac{17}{4}\Rightarrow 4{{x}^{2}}+4=17x\Rightarrow 4{{x}^{2}}-17x+4=0\Rightarrow 4{{x}^{2}}-16x-x+4=0\Rightarrow 4x\left( x-4 \right)-1\left( x-4 \right)=0\Rightarrow \left( x-4 \right)\left( 4x-1 \right)=0\therefore x=4\,or\,x=\dfrac{1}{4}$4 Marks Questions 1. Find the roots of the following Quadratic Equations by applying quadratic formulas. i).$2{{x}^{2}}-\text{ }7x+\text{ }3\text{ }=\text{ }0$Ans:$2{{x}^{2}}-\text{ }7x+\text{ }3\text{ }=\text{ }0$Comparing quadratic equation$2{{x}^{2}}-\text{ }7x+\text{ }3\text{ }=\text{ }0$with general form$a{{x}^{2}}+bx+c=0$, we get$a=2,\text{ }b=-7\text{ }and\text{ }c=3$Putting these values in quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}x=\dfrac{7\pm \sqrt{{{\left( -7 \right)}^{2}}-4\left( 2 \right)\left( 3 \right)}}{2\times 2}\Rightarrow x=\dfrac{7\pm \sqrt{49-24}}{4}\Rightarrow x=\dfrac{7\pm 5}{4}\Rightarrow x=\dfrac{7+5}{4},\dfrac{7-5}{4}\therefore x=3,\dfrac{1}{2}$ii).$2{{x}^{2}}+\text{ }x\text{ }\text{ }4\text{ }=\text{ }0$Ans: Comparing quadratic equation$2{{x}^{2}}+\text{ }x\text{ }\text{ }4\text{ }=\text{ }0$with the general form$a{{x}^{2}}+bx+c=0$, we get$a=2,\text{ }b=1\text{ }and\text{ }c=-4$Putting these values in quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} x=\dfrac{-1\pm \sqrt{{{1}^{2}}-4\left( 2 \right)\left( -4 \right)}}{2\times 2}  \Rightarrow x=\dfrac{-1\pm \sqrt{33}}{4}  \Rightarrow x=\dfrac{-1-\sqrt{33}}{4},\dfrac{-1+\sqrt{33}}{4} $iii).$4{{x}^{2}}+4\sqrt{3}x+3=0$Ans: Comparing quadratic equation$4{{x}^{2}}+4\sqrt{3}x+3=0$with the general form$a{{x}^{2}}+bx+c=0$, we get$a=4,\text{ }b=4\sqrt{3}\text{ }and\text{ }c=3x=\dfrac{-4\sqrt{3}\pm \sqrt{{{\left( 4\sqrt{3} \right)}^{2}}-4\left( 4 \right)\left( 3 \right)}}{2\times 4}\Rightarrow x=\dfrac{-4\sqrt{3}\pm \sqrt{0}}{8}\Rightarrow x=\dfrac{-\sqrt{3}}{2}$A quadratic equation has two roots. Here, both the roots are equal. Therefore,$x=\dfrac{-\sqrt{3}}{2},\dfrac{-\sqrt{3}}{2}$iv).$2{{x}^{2}}+\text{ }x\text{ }+\text{ }4\text{ }=\text{ }0$Ans:$2{{x}^{2}}+\text{ }x\text{ }+\text{ }4\text{ }=\text{ }0$Comparing quadratic equation$2{{x}^{2}}+\text{ }x\text{ }+\text{ }4\text{ }=\text{ }0$with the general form$a{{x}^{2}}+bx+c=0$, we get$a=2,b=1\text{ }and\text{ }c=\text{ }4$Putting these values in quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}x=\dfrac{-1\pm \sqrt{{{\left( 1 \right)}^{2}}-4\left( 2 \right)\left( 4 \right)}}{2\times 2}\Rightarrow x=\dfrac{-1\pm \sqrt{-31}}{4}$But, the square root of a negative numbers is not defined. Therefore, Quadratic Equation$2{{x}^{2}}+\text{ }x\text{ }+\text{ }4\text{ }=\text{ }0$has no solution. 2. An express train takes$~1$hour less than a passenger train to travel$132$km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If, the average speed of the express train is$11$km/h more than that of the passenger train, find the average speed of two trains Ans: Let average speed of passenger train$=\text{ }x$km/h Let average speed of express train$=\text{ }\left( x+11 \right)$km/h Time taken by passenger train to cover$132$km$=\dfrac{132}{x}$hours Time taken by express train to cover$132$km$=\left( \dfrac{132}{x+11} \right)$hours According to the given condition$\dfrac{132}{x}=\dfrac{132}{x+11}+1\Rightarrow 132\left( \dfrac{1}{x}-\dfrac{1}{x+11} \right)=1132\left( \dfrac{x+11-x}{x\left( +11 \right)} \right)=1\Rightarrow \text{ }132\left( 11 \right)=x\left( x+11 \right)\Rightarrow \text{ }1452={{x}^{2}}+11x\Rightarrow \text{ }{{x}^{2}}+11x\text{ }1452=0$Comparing equation${{x}^{2}}+11x\text{ }1452=0$with general quadratic equation$a{{x}^{2}}+bx+c=0$, we get$a=1,b=11\text{ }and\text{ }c=-1452$Applying Quadratic Formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}x=\dfrac{-11\pm \sqrt{{{\left( 11 \right)}^{2}}-4\left( 1 \right)\left( -1452 \right)}}{2\times 1}\Rightarrow x=\dfrac{-11\pm \sqrt{121+5808}}{2}\Rightarrow x=\dfrac{-11\pm \sqrt{5929}}{2}\Rightarrow x=\dfrac{-11\pm 77}{2}\Rightarrow x=\dfrac{-11+77}{2},\dfrac{-11-77}{2}\therefore x=33,-44$As speed cannot be negative. Therefore, speed of passenger train$=\text{ }33$km/h And, speed of express train$=\text{ }x+11=33+11=44$km/h 3. Sum of areas of two squares is$468\text{ }{{m}^{2}}$. If, the difference of their perimeters is$24$meters, find the sides of the two squares. Ans: Let perimeter of first square$=\text{ }x$meters Let perimeter of second square$=\text{ }\left( x+24 \right)$meters Length of side of first square$~=\dfrac{x}{4}$meters {Perimeter of square$=\text{ }4\text{ }\times $length of side} Length of side of second square$==\left( \dfrac{x+24}{4} \right)$meters Area of first square$= side \times side $=$\dfrac{x}{4}\times \dfrac{x}{4}=\dfrac{{{x}^{2}}}{16}{{m}^{2}}$Area of second square$={{\left( \dfrac{x+24}{4} \right)}^{2}}{{m}^{2}}$According to given condition:$\dfrac{{{x}^{2}}}{16}+{{\left( \dfrac{x+24}{4} \right)}^{2}}=468\Rightarrow \dfrac{{{x}^{2}}}{16}+\dfrac{{{x}^{2}}+576+48x}{16}=468\Rightarrow \dfrac{{{x}^{2}}+{{x}^{2}}+576+48x}{16}=468\Rightarrow \text{ }2{{x}^{2}}+576+48x=468\times 16\Rightarrow \text{ }2{{x}^{2}}+48x+576=7488\text{ }\Rightarrow \text{ }2{{x}^{2}}+48x\text{ }6912=0\text{ }\Rightarrow \text{ }{{x}^{2}}+24x\text{ }3456=0$Comparing equation${{x}^{2}}+24x\text{ }3456=0$with standard form$a{{x}^{2}}+bx+c=0$, We get$a=1,b=24\text{ }and\,c=\text{ }-3456$Applying Quadratic Formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}x=\dfrac{-24\pm \sqrt{{{\left( 24 \right)}^{2}}-4\left( 1 \right)\left( -3456 \right)}}{2\times 1}\Rightarrow x=\dfrac{-24\pm \sqrt{576+13824}}{2}\Rightarrow x=\dfrac{-24\pm \sqrt{14400}}{2}=\dfrac{-24\pm 120}{2}\Rightarrow x=\dfrac{-24+120}{2},\dfrac{-24-120}{2}\therefore x=48,-72$Perimeter of square cannot be in negative. Therefore, we discard$x=-72$. Therefore, perimeter of first square$=\text{ }48$meters And, Perimeter of second square$=\text{ }x+24=48+24=72$meters$\Rightarrow $Side of First square$=\dfrac{Perimeter}{4}=\dfrac{48}{4}=12m$And, Side of second Square$=\dfrac{Perimeter}{4}=\dfrac{72}{4}=18m$4. Is it possible to design a rectangular park of perimeter$80$meters and area$400\text{ }{{m}^{2}}$. If so, find its length and breadth. Ans: Let length of park$=\text{ }x$meters We are given area of rectangular park$=\text{ }400\text{ }{{m}^{2}}$Therefore, breadth of park$=\dfrac{400}{x}$meters {Area of rectangle$~=$length$\times $breadth} Perimeter of rectangular park$=\text{ }2$(length$+$breath)$=\left( x+\dfrac{400}{x} \right)$meters We are given perimeter of rectangle$=\text{ }80$meters According to condition:$2\left( x+\dfrac{400}{x} \right)=80\Rightarrow 2\left( \dfrac{{{x}^{2}}+400}{x} \right)=80\Rightarrow \text{ }2{{x}^{2}}+800=80x\Rightarrow \text{ }2{{x}^{2}}-80x+800=0\Rightarrow \text{ }{{x}^{2}}-40x+400=0$Comparing equation,${{x}^{2}}-40x+400=0$with general quadratic equation$a{{x}^{2}}+bx+c=0$, we get$a=1,b=-40\text{ }and\text{ }c=400$Discriminant$=\text{ }{{b}^{2}}-4ac={{\left( -40 \right)}^{2}}\text{ }4\left( 1 \right)\left( 400 \right)=1600\text{ }\text{ }1600=0$Discriminant is equal to$0$. Therefore, two roots of equation are real and equal which means that it is possible to design a rectangular park of perimeter$80$meters and area$400\text{ }{{m}^{2}}$. Using quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$to solve equation,s$x=\dfrac{40\pm \sqrt{0}}{2}=\dfrac{42}{2}=20$Here, both the roots are equal to$20$. Therefore, length of rectangular park$=\text{ }20$meters Breadth of rectangular park$=\dfrac{400}{x}=\dfrac{400}{20}=20m$5. If I had walked$1$km per hour faster, I would have taken$10$minutes less to walk$2$km. Find the rate of my walking. Ans: Distance$=\text{ }2$km Let speed$=x$km/hr. New speed$~=\text{ }\left( x+1 \right)$km/hr. Time taken by normal speed$=\dfrac{2}{x}\,hr$Time taken by new speed =$\dfrac{2}{x+1}hr$According to question,$\dfrac{2}{x}-\dfrac{2}{x+1}=\dfrac{10}{60}\Rightarrow \dfrac{2x+2-2x}{{{x}^{2}}+x}=\dfrac{1}{6}\Rightarrow {{x}^{2}}+x=12\Rightarrow {{x}^{2}}+x-12=0\Rightarrow {{x}^{2}}+4x-3x-12=0\Rightarrow x\left( x+4 \right)-3\left( x+4 \right)=0\Rightarrow \left( x+4 \right)\left( x-3 \right)=0\therefore x=-4\,or\,x=3$So, speed is$x=3$km/hr 6. A takes$6$days less than the time taken by B to finish a piece of work. If both A and B together can finish it in$4$days, find the time taken by B to finish the work. Ans: Let B takes$x$days to finish the work, then A alone can finish it in$\left( x-6 \right)$days According to question,$\dfrac{1}{x}+\dfrac{1}{x-6}=\dfrac{1}{4}\Rightarrow \dfrac{x-6+x}{{{x}^{2}}-6x}=\dfrac{1}{4}\Rightarrow \dfrac{2x-6}{{{x}^{2}}-6x}=\dfrac{1}{4}\Rightarrow {{x}^{2}}-6x=8x-24\Rightarrow {{x}^{2}}-14x+24=0\Rightarrow {{x}^{2}}-12x-2x+24=0=x\left( x-12 \right)-2\left( x-12 \right)=0\Rightarrow \left( x-12 \right)\left( x-2 \right)=0\therefore x=12\,or\,x=2x=2$(Neglect) So, B takes$x=12$days. 7. A plane left$30$minutes later than the schedule time and in order to reach its$59$destination$1500$km away in time it has to increase its speed by$250$km/hr from its usual speed. Find its usual speed. Ans: Let usual speed$=x$km/hr New speed$=\left( x+250 \right)$km/hr Total distance$=\text{ }1500$km Time taken by usual speed$=\dfrac{1500}{x}$hr Time taken by new speed$=\dfrac{1500}{x+250}$hr According to question,$\dfrac{1500}{x}-\dfrac{1500}{x+250}=\dfrac{1}{2}\Rightarrow \dfrac{1500x+1500\times 250-1500x}{{{x}^{2}}+250x}=\dfrac{1}{2}\Rightarrow {{x}^{2}}+250x=\dfrac{1500\times 250}{2}\Rightarrow {{x}^{2}}+250x=750000\Rightarrow {{x}^{2}}+250x-750000=0\Rightarrow x^{2}+1000x-750x-750000=0\Rightarrow x\left( x+1000 \right)-750\left( x+1000 \right)=0\Rightarrow x=750\,\,or\,x=-1000$Therefore, usual speed is$750$km/hr,$-1000$is neglected as speed cannot be negative. 8. A motor boat, whose speed is$15$km/hr in still water, goes$30$km downstream and comes back in a total time of$4$hr$30$minutes, find the speed of the stream. Ans: Speed of motor boat in still water$=\text{ }15$km/hr Speed of stream$=x$km/hr Speed in downward direction$15+x$Speed in downward direction$15-x$According to question,$\dfrac{30}{15+x}+\dfrac{30}{15-x}=4\dfrac{1}{2}\Rightarrow \dfrac{30\left( 15-x \right)+30\left( 15+x \right)}{\left( 15+x \right)\left( 15-x \right)}=\dfrac{9}{2}\Rightarrow \dfrac{450-30x+450+30x}{225-{{x}^{2}}}=\dfrac{9}{2}\Rightarrow 9\left( 225-{{x}^{2}} \right)=1800\Rightarrow 225-{{x}^{2}}=200\therefore x=5$Speed of stream$~=\text{ }5$km/hr. 9. A swimming pool is filled with three pipes with uniform flow. The first two pipes operating simultaneously fill the pool in the same time during which the pool is the same time during which the pool is filled by the third pipe alone. The second pipe fills the pool five hours faster than the first pipe and four hours slower than the third pipe. Find the time required by each pipe to fill the pool separately. Ans: Let$x$be the number of hours required by the second pipe alone to till the pool and first pipe$\left( x+5 \right)$hour while third pipe$\left( x-4 \right)$hour$\dfrac{1}{x+5}+\dfrac{1}{x}=\dfrac{1}{x-4}\Rightarrow \dfrac{x+x+5}{{{x}^{2}}+5x}=\dfrac{1}{x-4}\Rightarrow {{x}^{2}}-8x-20=0\Rightarrow {{x}^{2}}-10x+2x-20=0\Rightarrow x\left( x-10 \right)+2\left( x-10 \right)=0\Rightarrow \left( x-10 \right)\left( x+2 \right)=0\Rightarrow x=10\,or\,x=-2\left( Neglected \right)$10. A two-digit number is such that the product of its digits is$18$. When$63$is subtracted from the number the digit interchanges their places. Find the number Ans: Let digit on unit’s place$=x$Digit on ten’s place$=yxy=18$(given) Number = 10.$y+x=10\left( \dfrac{18}{x} \right)+x$According to question,$10\left( \dfrac{18}{x} \right)+x-63=10x+\dfrac{18}{x}\Rightarrow \dfrac{180}{x}+\dfrac{x-63}{1}=\dfrac{10{{x}^{2}}+18}{x}\Rightarrow \dfrac{180+{{x}^{2}}-63x}{x}=\dfrac{10{{x}^{2}}+18}{x}\Rightarrow 9{{x}^{2}}+63x-162=0\Rightarrow 9\left( {{x}^{2}}+9x-2x-18 \right)=0\Rightarrow x\left( x+9 \right)-2\left( x+9 \right)=0\Rightarrow x=2\,or\,x=-9$Number$=10\left( \dfrac{18}{2} \right)+2=92$11. A factory kept increasing its output by the same percent ago every year. Find the percentage if it is known that the output is doubled in the last two years. Ans: According to question, 2P$=p{{\left( 1+\dfrac{r}{100} \right)}^{2}}\Rightarrow \dfrac{\sqrt{2}}{1}=1+\dfrac{r}{100}\sqrt{2}-1=\dfrac{r}{100}\Rightarrow r=\left( \sqrt{2}-1 \right)100$12. Two pipes running together can fill a cistern in if one pipe takes$3\dfrac{1}{13}$minutes more than the other to fill it, find the time in which each pipe would fill the cistern. Ans: Let the faster pipe takes minutes to fill the cistern and the slower pipe will take$\left( x+3 \right)$minutes. According to question,$\dfrac{1}{x}+\dfrac{1}{x+3}=\dfrac{1}{\dfrac{40}{13}}\Rightarrow \dfrac{1}{x}+\dfrac{1}{x+3}=\dfrac{13}{40}\Rightarrow \dfrac{x+3+x}{{{x}^{2}}+3x}=\dfrac{13}{40}\Rightarrow 13{{x}^{2}}-41x-120=0\Rightarrow 13{{x}^{2}}-65x+24x-120=0\Rightarrow 13x\left( x-5 \right)+24\left( x-5 \right)=0\therefore x=5\,or\,x=\dfrac{-24}{13}\left( Neglected \right)$13. If the roots of the equation$\left( a-b \right){{x}^{2}}+\left( b-c \right)x+\left( c-a \right)=0$are equal, prove that$2a=b+c$Ans:$\left( a-b \right){{x}^{2}}+\left( b-c \right)x+\left( c-a \right)=0D={{b}^{2}}-4ac={{\left( b-c \right)}^{2}}-4\times \left( a-b \right)\times \left( c-a \right)={{b}^{2}}+{{c}^{2}}-2bc-4\left( ac-{{a}^{2}}-bc+ab \right)={{b}^{2}}+{{c}^{2}}-2bc-4ac+4{{a}^{2}}+4bc-4ab={{\left( b \right)}^{2}}+{{\left( c \right)}^{2}}+{{\left( 2a \right)}^{2}}+2bc-4ac-4ab={{\left( b+c-2a \right)}^{2}}$For equal root s, D = 0$\Rightarrow {{\left( b+c-2a \right)}^{2}}=0\Rightarrow \left( b+c-2a \right)=0\Rightarrow b+c=2a$14. Two circles touch internally. The sum of their areas is$116\pi c{{m}^{2}}$and the distance between their centers is$6$cm. Find the radii of the circles. Ans: Let${{r}_{1}}$and${{r}_{2}}$be the radius of two circles According to question,$\Pi {{r}_{1}}^{2}+\Pi {{r}_{2}}^{2}=116\Pi \Rightarrow {{r}_{1}}^{2}+{{r}_{2}}^{2}=116......\left( i \right){{r}_{2}}-{{r}_{1}}=6$(Given)$\Rightarrow {{r}_{2}}=6+{{r}_{1}}$Puting the value of${{r}_{2}}$in eq. … (i) we get$\Rightarrow {{r}_{1}}^{2}+36+{{r}_{1}}^{2}+12{{r}_{1}}=116\Rightarrow 2{{r}_{1}}^{2}+12{{r}_{1}}-80=0{{r}_{1}}^{2}+6{{r}_{1}}-40=0\Rightarrow {{r}_{1}}^{2}+10{{r}_{1}}-4{{r}_{1}}-40=0\Rightarrow {{r}_{1}}\left( {{r}_{1}}+10 \right)-4\left( {{r}_{1}}+10 \right)=0\Rightarrow \left( {{r}_{1}}+10 \right)\left( {{r}_{1}}-4 \right)=0when\,{{r}_{1}}=4\,cm{{r}_{2}}=6+{{r}_{1}}=6+4{{r}_{2}}=10\,cm\Rightarrow {{r}_{1}}=-10$(Neglect) or${{r}_{1}}=\text{ }4\text{ }cm$15. A piece of cloth costs Rs.$200$. If the piece was$5$m longer and each metre of cloth costs Rs.$2$less the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per meter? Ans: Let the length of piece$=x$m Rate per meter$=\dfrac{200}{x}$A New length$=\text{ }\left( x+5 \right)$A New rate per meter$=\dfrac{200}{x+5}$According to question,$\dfrac{200}{x}-\dfrac{200}{x+5}=2\Rightarrow \dfrac{200\left( x+5 \right)-200x}{\left( x+5 \right)}=\dfrac{2}{1}\Rightarrow \dfrac{200x+1000-200x}{{{x}^{2}}+5x}=\dfrac{2}{1}\Rightarrow {{x}^{2}}+5x=500\Rightarrow {{x}^{2}}+25x-20x-500=0\Rightarrow x\left( x+25 \right)-20\left( x+25 \right)=0\Rightarrow \left( x+25 \right)\left( x-20 \right)=0\therefore x=-25\left( Neglect \right)\,or\,x=20$Rate per meter$=\text{ }10$16.$a{{x}^{2}}+bx+x\text{ }=\text{ }0$,$a\ne 0$solve by quadratic formula. Ans:$a{{x}^{2}}+bx+x\text{ }=\text{ }0\Rightarrow {{x}^{2}}+\dfrac{b}{a}x+\dfrac{c}{a}=0\Rightarrow {{x}^{2}}+\dfrac{b}{a}x{{\left( \dfrac{b}{2a} \right)}^{2}}-{{\left( \dfrac{b}{2a} \right)}^{2}}+\dfrac{c}{a}=0\Rightarrow {{\left( x+\dfrac{b}{2a} \right)}^{2}}=\dfrac{{{b}^{2}}}{4{{a}^{2}}}-\dfrac{c}{a}$x$\Rightarrow {{\left( x+\dfrac{b}{2a} \right)}^{2}}=\dfrac{{{b}^{2}}-4ac}{4{{a}^{2}}}\Rightarrow x+\dfrac{b}{2a}=\pm \sqrt{\dfrac{{{b}^{2}}-4ac}{4{{a}^{2}}}}\Rightarrow x+\dfrac{-b}{2a}=\pm \sqrt{\dfrac{{{b}^{2}}-4ac}{2a}}\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\Rightarrow x=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}or\,\Rightarrow x=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$17. The length of the hypotenuse of a right-angled triangle exceeds the length of the base by$2$cm and exceeds twice the length of the altitude by$1$cm. Find the length of each side of the triangle. Ans: Let base of the triangle$=\text{ }x$Altitude of the triangle$=\text{ }y$Hypotenuse of the triangle$=\text{ }h$According to question,$h=x+2h=2y+1\Rightarrow x+2=2y+1\Rightarrow x+2-1=2y\Rightarrow x-1=2y\Rightarrow \dfrac{x-1}{2}=yAnd\,{{x}^{2}}+{{y}^{2}}={{h}^{2}}\Rightarrow {{x}^{2}}+{{\left( \dfrac{x-1}{2} \right)}^{2}}={{\left( x+2 \right)}^{2}}\Rightarrow {{x}^{2}}-15x+x-15=0\Rightarrow \left( x-15 \right)\left( x+1 \right)=0\Rightarrow x=15\,or\,x=-1$Base of the triangle$=\text{ }15$cm Altitude of the triangle$=\dfrac{x+1}{2}=8cm$Hypotenuse of the triangle$=\text{ }17$cm 18. Find the roots of the following quadratic equations if they exist by the method of completing square. i).$2{{x}^{2}}\text{ }-7x+3=0$Ans: (i)$2{{x}^{2}}\text{ }-7x+3=0$Dividing the equation by$2$to make coefficient of${{x}^{2}}$equal to$1$we get${{x}^{2}}-\dfrac{7}{2}x+\dfrac{3}{2}=0$Dividing the middle term of the equation by$2x$, we get$\dfrac{7}{2}x\times \dfrac{1}{2x}=\dfrac{7}{4}$Adding and subtracting square of$\dfrac{7}{4}$from the equation${{x}^{2}}-\dfrac{7}{2}x+\dfrac{3}{2}=0$we get${{x}^{2}}-\dfrac{7}{2}x+\dfrac{3}{2}+{{\left( \dfrac{7}{4} \right)}^{2}}-{{\left( \dfrac{7}{4} \right)}^{2}}=0\Rightarrow {{x}^{2}}+{{\left( \dfrac{7}{4} \right)}^{2}}-\dfrac{7}{2}x+\dfrac{3}{2}+-{{\left( \dfrac{7}{4} \right)}^{2}}=0\left\{ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \right\}\Rightarrow {{\left( \text{x-}\dfrac{\text{7}}{\text{4}} \right)}^{\text{2}}}\text{+}\dfrac{\text{24-49}}{\text{16}}\text{=0}\Rightarrow {{\left( \text{x-}\dfrac{\text{7}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{49-24}}{\text{16}}$Square rooting on both the sides we get$\Rightarrow \text{x-}\dfrac{\text{7}}{\text{4}}\text{= }\!\!\pm\!\!\text{ }\dfrac{\text{5}}{\text{4}}\Rightarrow x\text{=}\dfrac{\text{5}}{\text{4}}\text{+}\dfrac{\text{7}}{\text{4}}\text{=}\dfrac{\text{12}}{\text{4}}\text{=3}\,\text{and}\,\text{x=-}\dfrac{\text{5}}{\text{4}}\text{+}\dfrac{\text{7}}{\text{4}}\text{=}\dfrac{\text{2}}{\text{4}}\text{=}\dfrac{\text{1}}{\text{2}}$Therefore,$x=\dfrac{1}{2},3$ii).$2{{x}^{2}}+x\text{ }4=0$Ans:$\text{2}{{\text{x}}^{\text{2}}}\text{+x-- 4=0}$Dividing equation by$2$,${{\text{x}}^{\text{2}}}\text{+}\dfrac{\text{x}}{\text{2}}\text{-2=0}$Following procedure of completing square,${{\text{x}}^{\text{2}}}\text{+}\dfrac{\text{x}}{\text{2}}\text{-2+}{{\left( \dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{-}{{\left( \dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{=0}\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\dfrac{\text{x}}{\text{2}}\text{+}{{\left( \dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{-2-}\dfrac{\text{1}}{\text{16}}\text{=0}\left\{ {{\left( \text{a-b} \right)}^{\text{2}}}\text{=}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{-2ab} \right\}\Rightarrow {{\left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{-}\dfrac{\text{33}}{\text{16}}\text{=0}\Rightarrow {{\left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{-}\dfrac{\text{33}}{\text{16}}$Taking square root on both sides,$\Rightarrow \text{x+}\dfrac{\text{1}}{\text{4}}\text{= }\!\!\pm\!\!\text{ }\dfrac{\sqrt{\text{33}}}{\text{4}}\Rightarrow x=\dfrac{\sqrt{33}}{4}-\dfrac{1}{4}\text{=}\dfrac{\sqrt{\text{33}}\text{-1}}{\text{4}}\,\text{and}\,\text{x=-}\dfrac{\sqrt{\text{33}}}{\text{4}}\text{-}\dfrac{\text{1}}{\text{4}}\text{=}\dfrac{\text{-}\sqrt{\text{33}}\text{-1}}{\text{4}}$Therefore,$\text{x}=\dfrac{\sqrt{33}-1}{4},\dfrac{-\sqrt{33}-1}{4}$iii).$4{{x}^{2}}+4\sqrt{3}x+3=0$Ans:$\text{4}{{\text{x}}^{\text{2}}}\text{+4}\sqrt{\text{3}}\text{x+3=0}$Dividing this equation by$4,$we get${{\text{x}}^{\text{2}}}\text{+}\sqrt{\text{3}}\text{x+}\dfrac{\text{3}}{\text{4}}\text{=0}$By the procedure of completing square we get$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\sqrt{\text{3}}\text{x+}\dfrac{\text{3}}{\text{4}}\text{+}{{\left( \dfrac{\sqrt{\text{3}}}{\text{2}} \right)}^{\text{2}}}\text{-}{{\left( \dfrac{\sqrt{\text{3}}}{\text{2}} \right)}^{\text{2}}}\text{=0}\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\left( \dfrac{\sqrt{\text{3}}}{\text{2}} \right)}^{\text{2}}}\text{+}\sqrt{\text{3}}\text{x+}\dfrac{\text{3}}{\text{4}}\text{-}\dfrac{\text{3}}{\text{4}}\text{=0}\left\{ {{\left( \text{a-b} \right)}^{\text{2}}}\text{=}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{-2ab} \right\}\Rightarrow {{\left( \text{x+}\dfrac{\sqrt{\text{3}}}{\text{2}} \right)}^{\text{2}}}\text{=0}\Rightarrow \left( \text{x+}\dfrac{\sqrt{\text{3}}}{\text{2}} \right)\left( \text{x+}\dfrac{\sqrt{\text{3}}}{\text{2}} \right)\text{=0}\Rightarrow \text{x+}\dfrac{\sqrt{\text{3}}}{\text{2}}\text{=0,x+}\dfrac{\sqrt{\text{3}}}{\text{2}}\text{=0}\Rightarrow \text{x=-}\dfrac{\sqrt{\text{3}}}{\text{2}}\text{,-}\dfrac{\sqrt{\text{3}}}{\text{2}}$iv).$2{{x}^{2}}+x+4=0$Ans:$\text{2}{{\text{x}}^{\text{2}}}\text{+x+4=0}$Dividing this equation by$2$we get${{\text{x}}^{\text{2}}}\text{+}\dfrac{\text{x}}{\text{2}}\text{+2=0}$By the procedure of completing square,$\Rightarrow {{x}^{2}}+\dfrac{x}{2}+2+{{\left( \dfrac{1}{4} \right)}^{2}}-{{\left( \dfrac{1}{4} \right)}^{2}}=0\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\left( \dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{+}\dfrac{\text{x}}{\text{2}}\text{+2-}{{\left( \dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{=0}\left\{ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \right\}\Rightarrow {{\left( x+\dfrac{1}{4} \right)}^{2}}+2-\dfrac{1}{16}=0{{\left( x+\dfrac{1}{4} \right)}^{2}}=\dfrac{1}{16}-2=\dfrac{1-32}{16}$Right hand side does not exist because the square root of the negative number does not exist. Therefore, there is no solution for quadratic equation$2{{x}^{2}}+x+4=0\$

### Practice Questions for Class 10 Maths Chapter 4 - Quadratic Equations

The following are some of the questions that can be taken up by students to assist them in the board preparations related to Quadratic Equations.

Question 1. A rectangular field's diagonal is 60 metres longer than the shorter side. Find the field's sides if the longer side is 30 metres more than the shorter side.

Anwer- 120m which is the correct answer.

Question 2. Is it possible to build a rectangular park with an 80 meter perimeter and a 400-square-meter area? If this is the case, determine its length and breadth.

Question 3- A train moving at a speed of 600 km was slowed down due to bad weather. The train's average speed dropped by 200 km/hr, and the journey time was increased by 30 minutes. Determine the train's initial duration.

Conclusion

Vedantu's goal is to help students from all over the country prepare for exams. As a result, all of our study materials are available in PDF format, which can be downloaded for free. Our professionals answer the questions to provide students with a sample answer for each question on the Class 10 CBSE test papers. For practice, students can download and solve the question paper on their own. For exam preparation, they can use Vedantu to access critical questions, revision notes, NCERT chapter-by-chapter solutions, and other resources.

## FAQs on Important Questions for CBSE Class 10 Maths Chapter 4 - Quadratic Equations

Q1: What are the roots of a quadratic equation? Explain the nature of roots of a quadratic equation.

A1: The roots of the quadratic equation are given as :  x = (-b ± √D)/2a, where D = b2 – 4ac.

The nature of  the roots of a quadratic equation are as mentioned below:

• D > 0, roots are real and distinct (unequal)

• D = 0, roots are real and equal (coincident)

• D < 0, roots are imaginary and unequal

Q2: Explain in detail the nature of roots of a quadratic equation.

A2: The nature of roots of a quadratic equation is as follows:

• If the value of discriminant = 0 i.e. b2 – 4ac = 0, then the quadratic equation will have equal roots i.e. α = β = -b/2a.

• If the value of discriminant < 0 i.e. b2 – 4ac < 0, then the quadratic equation will have imaginary roots i.e α = (p + iq) and β = (p – iq). Where ‘iq’ is the imaginary part of a complex number.

• If the value of discriminant (D) > 0 i.e. b2 – 4ac > 0, then the quadratic equation will have real roots.

• If the value of discriminant > 0 and D is a perfect square, then the quadratic equation will have rational roots.

• If the value of discriminant (D) > 0 and D is not a perfect square, then the quadratic equation will have irrational roots i.e. α = (p + √q) and β=(p – √q).

• If the value of discriminant > 0, D is a perfect square, a = 1 and b and c are integers, then the quadratic equation will have integral roots.

Q3: What is the Relationship between Coefficient and Roots of  Quadratic Equation?

A3: If α and β are roots of a Quadratic Equation ax2 + bx + c then,

• α + β = -b/a

• αβ = c/a

• α – β = ±√[(α + β)2 – 4αβ]

• |α + β| = √D/|a|

The relationship between the roots and coefficient of a polynomial equation can be derived by simplifying the given polynomials and substituting the above results as shown below.

• α2β + β2α = αβ (α + β) = – bc/a2

• α2 + αβ + β2 = (α + β)2 – αβ = (b2 – ac)/a2

• α2 + β2 = (α – β)2 – 2αβ

• α2 – β2 = (α + β) (α – β)

• α3 + β3 = (α + β)3 + 3αβ(α + β)

• α3 – β3 = (α – β)3 + 3αβ(α – β)

• (α/β)2 + (β/α)2 = α4 + β42β2

Q4: What happens when a < 0 and  b2 – 4ac > 0.

A4: When a < 0 and  b2 – 4ac > 0, the graph of a quadratic equation will be concave downwards and will intersect the x-axis at two points α and β with α < β. The quadratic equation will have two real roots (α and β) and the curve will always lie below the x-axis.

• The quadratic function f(x) will be positive i.e. f(x) > 0, for the values of x lying in the interval (α, β).

• The quadratic function f(x) will be equal to zero i.e. f(x) = 0, if x = α or β

• The quadratic function f(x) will be negative i.e. f(x) < 0 for the values of x lying in the interval (−∞, α) ∪ (β, ∞).