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NCERT Solutions for Class 10 Maths Chapter 4: Quadratic Equations - Exercise 4.3

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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Free PDF download of NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3 (Ex 4.3) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all NCERT Book Solutions in your emails.


Class:

NCERT Solutions for Class 10

Subject:

Class 10 Maths

Chapter Name:

Chapter 4 - Quadratic Equations

Exercise:

Exercise - 4.3

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes


 

You can also Download Class 10 Science NCERT Solutions along with Maths NCERT Solutions Class 10 to help you to revise complete Syllabus and score more marks in your examinations.

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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.3

Topics to Learn before Solving Exercise 4.3

Steps for finding Roots of an equation by using Quadratic Formula

  • First, we have to write the equation in the standard form.

  • Then we will compare the equation with ax2 + bx + c = 0 and will find the values of a, b, and c.

  • We will substitute the values in the quadratic formula, x = [-b ± √(b² - 4ac)] / (2a)].

  • Then we will simplify the equation so obtained.


Steps in Solving Quadratic Equations by Completing Square

  • First, we have to write the equation in the standard form.

  • Then we will complete the square on LHS. 

  • Then we will simplify. 


Points to Remember:

  • Using the completing square method and the quadratic formula method, you can solve any sort of quadratic equation.

  • The roots of a quadratic equation are also referred to as "solutions" or "zeros".

  • For any quadratic equation ax2 + bx + c = 0,

  • Sum of the roots = -b/a

  • Product of the roots = c/a.


Exercise 4.3

1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

i. $\text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=0}$

Ans. $\text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=0}$

On dividing both sides of the equation by $\text{2}$.

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-}\dfrac{\text{7}}{\text{2}}\text{x=-}\dfrac{\text{3}}{\text{2}}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-2}\left( \dfrac{\text{7}}{\text{4}} \right)\text{x=-}\dfrac{\text{3}}{\text{2}}$

On adding ${{\left( \dfrac{7}{4} \right)}^{2}}$ both sides of equation.

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-2}\left( \dfrac{\text{7}}{\text{4}} \right)\text{x+}{{\left( \dfrac{7}{4} \right)}^{2}}\text{=-}\dfrac{\text{3}}{\text{2}}+{{\left( \dfrac{7}{4} \right)}^{2}}$

$\Rightarrow {{\left( \text{x-}\dfrac{\text{7}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{49}}{\text{16}}\text{-}\dfrac{\text{3}}{\text{2}}$

$\Rightarrow {{\left( \text{x-}\dfrac{\text{7}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{25}}{\text{16}}$

$\Rightarrow \left( \text{x-}\dfrac{\text{7}}{\text{4}} \right)\text{=}\pm \dfrac{5}{4}$

$\Rightarrow x\text{=}\dfrac{\text{7}}{\text{4}}\pm \dfrac{5}{4}$

$\Rightarrow x\text{=}\dfrac{\text{7}}{\text{4}}+\dfrac{5}{4}$ or $x\text{=}\dfrac{\text{7}}{\text{4}}-\dfrac{5}{4}$

$\Rightarrow x\text{=}\dfrac{12}{\text{4}}$ or $x\text{=}\dfrac{2}{\text{4}}$

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

 

ii. $\text{2}{{\text{x}}^{\text{2}}}\text{+x-4=0}$

Ans. $\text{2}{{\text{x}}^{\text{2}}}\text{+x-4=0}$

On dividing both sides of the equation by $\text{2}$.

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\dfrac{1}{2}\text{x=2}$

On adding ${{\left( \dfrac{1}{4} \right)}^{2}}$ both sides of equation.

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

$\Rightarrow {{\left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{33}}{\text{16}}$

\[\Rightarrow \left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)\text{=}\pm \dfrac{\sqrt{\text{33}}}{4}\]

\[\Rightarrow \text{x= }\!\!\pm\!\!\text{ }\dfrac{\sqrt{\text{33}}}{\text{4}}\text{-}\dfrac{\text{1}}{\text{4}}\]

\[\Rightarrow \text{x= }\!\!\pm\!\!\text{ }\dfrac{\sqrt{\text{33}}-1}{\text{4}}\]

\[\Rightarrow \text{x=}\dfrac{\sqrt{\text{33}}-1}{\text{4}}\] or \[\Rightarrow \text{x=}\dfrac{-\sqrt{\text{33}}-1}{\text{4}}\]

 

iii. $\text{4}{{\text{x}}^{\text{2}}}\text{+4}\sqrt{3}\text{x+3=0}$

Ans. $\text{4}{{\text{x}}^{\text{2}}}\text{+4}\sqrt{3}\text{x+3=0}$

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

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

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

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

iv. $\text{2}{{\text{x}}^{\text{2}}}\text{+x+4=0}$

Ans. $\text{2}{{\text{x}}^{\text{2}}}\text{+x+4=0}$

On dividing both sides of the equation by $\text{2}$.

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\dfrac{\text{1}}{\text{2}}\text{x+2=0}$

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

On adding ${{\left( \dfrac{1}{4} \right)}^{2}}$ both sides of equation.

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

$\Rightarrow {{\left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{1}}{\text{16}}\text{-2}$

$\Rightarrow {{\left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{=-}\dfrac{\text{31}}{\text{16}}$

Since, the square of a number cannot be negative.

Therefore, there is no real root for the given equation.

 

2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

i. $\text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=0}$

Ans. $\text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=2}$, $\text{b=-7}$, $\text{c=3}$.

Therefore, by using quadratic formula-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

$\Rightarrow \text{x=}\dfrac{\text{7 }\!\!\pm\!\!\text{ }\sqrt{\text{49-24}}}{4}$

$\Rightarrow \text{x=}\dfrac{\text{7 }\!\!\pm\!\!\text{ 5}}{4}$

$\Rightarrow \text{x=}\dfrac{\text{7+5}}{4}$  or $\Rightarrow \text{x=}\dfrac{\text{7-5}}{4}$

$\Rightarrow \text{x=}\dfrac{12}{4}$  or $\Rightarrow \text{x=}\dfrac{\text{2}}{4}$

$\therefore \text{x=3 or }\dfrac{\text{1}}{\text{2}}$.

 

ii. $\text{2}{{\text{x}}^{\text{2}}}\text{+x-4=0}$

Ans. $\text{2}{{\text{x}}^{\text{2}}}\text{+x-4=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=2}$, $\text{b=1}$, $\text{c=-4}$.

Therefore, by using quadratic formula-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

$\Rightarrow \text{x=}\dfrac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{\text{1-32}}}{4}$

$\Rightarrow \text{x=}\dfrac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{33}}{4}$

$\Rightarrow \text{x=}\dfrac{\text{-1+}\sqrt{33}}{4}$  or $\Rightarrow \text{x=}\dfrac{\text{-1-}\sqrt{33}}{4}$

$\therefore \text{x=}\dfrac{\text{-1+}\sqrt{33}}{4}\text{ or }\dfrac{\text{-1-}\sqrt{33}}{4}$.

 

iii.$\text{4}{{\text{x}}^{\text{2}}}\text{+4}\sqrt{3}\text{x+3=0}$

Ans. $\text{4}{{\text{x}}^{\text{2}}}\text{+4}\sqrt{3}\text{x+3=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=4}$, $\text{b=4}\sqrt{3}$, $\text{c=3}$.

Therefore, by using quadratic formula-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

$\Rightarrow \text{x=}\dfrac{\text{-4}\sqrt{3}\text{ }\!\!\pm\!\!\text{ }\sqrt{\text{48-48}}}{8}$

$\Rightarrow \text{x=}\dfrac{\text{-4}\sqrt{3}}{8}$

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

$\therefore \text{x=}\dfrac{\text{-}\sqrt{3}}{2}\text{ or }\dfrac{\text{-}\sqrt{3}}{2}$.

 

iv. $\text{2}{{\text{x}}^{\text{2}}}\text{+x+4=0}$

Ans. $\text{2}{{\text{x}}^{\text{2}}}\text{+x+4=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=2}$, $\text{b=1}$, $\text{c=4}$.

Therefore, by using quadratic formula-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

\[\Rightarrow \text{x=}\dfrac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{\text{1-32}}}{4}\]

$\Rightarrow \text{x=}\dfrac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{\text{-31}}}{4}$

Since, the square of a number cannot be negative.

Therefore, there is no real root for the given equation.

 

3.Find the roots of the following equations:

i. $\text{x-}\dfrac{\text{1}}{\text{x}}\text{=3,x}\ne \text{0}$

Ans: $\text{x-}\dfrac{\text{1}}{\text{x}}\text{=3,x}\ne \text{0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=1}$, $\text{b=-3}$, $\text{c=-1}$.

Therefore, by using quadratic formula-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

$\Rightarrow \text{x=}\dfrac{\text{3 }\!\!\pm\!\!\text{ }\sqrt{\text{9+4}}}{2}$

$\Rightarrow \text{x=}\dfrac{\text{3 }\!\!\pm\!\!\text{ }\sqrt{\text{13}}}{2}$

$\Rightarrow \text{x=}\dfrac{\text{3+}\sqrt{\text{13}}}{2}$  or $\Rightarrow \text{x=}\dfrac{\text{3-}\sqrt{\text{13}}}{2}$

$\therefore \text{x=}\dfrac{\text{3+}\sqrt{\text{13}}}{2}\text{ or }\dfrac{\text{3-}\sqrt{\text{13}}}{2}$.

 

ii.$\dfrac{\text{1}}{\text{x+4}}\text{-}\dfrac{\text{1}}{\text{x-7}}\text{=}\dfrac{\text{11}}{\text{30}}\text{,x}\ne \text{-4,7}$

Ans:$\dfrac{\text{1}}{\text{x+4}}\text{-}\dfrac{\text{1}}{\text{x-7}}\text{=}\dfrac{\text{11}}{\text{30}}\text{,x}\ne \text{-4,7}$

$\Rightarrow \dfrac{\text{x-7-x-4}}{\left( \text{x+4} \right)\left( \text{x-7} \right)}\text{=}\dfrac{\text{11}}{\text{30}}$

$\Rightarrow \dfrac{\text{-11}}{\left( \text{x+4} \right)\left( \text{x-7} \right)}\text{=}\dfrac{\text{11}}{\text{30}}$

$\Rightarrow \left( \text{x+4} \right)\left( \text{x-7} \right)=-30$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-3x-28=-3}0$

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

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

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

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

$\Rightarrow x\text{=1 or 2}$

 

4. The sum of the reciprocals of Rehman’s ages, (in years) $\text{3}$ years ago and $\text{5}$ years from now is \[\dfrac{\text{1}}{\text{3}}\]. Find his present age.

Ans. Let the present age of Rehman be $\text{x}$ years.

Three years ago, his age is $\left( \text{x-3} \right)\text{years}$.

Five years hence, his age will be $\left( \text{x+5} \right)\text{years}$.

Therefore,$\dfrac{\text{1}}{\text{x-3}}\text{+}\dfrac{\text{1}}{\text{x+5}}\text{=}\dfrac{\text{1}}{\text{3}}$

\[\Rightarrow \dfrac{\text{x+5+x-3}}{\left( \text{x-3} \right)\left( \text{x+5} \right)}\text{=}\dfrac{\text{1}}{\text{3}}\]

\[\Rightarrow \dfrac{\text{2x+2}}{\left( \text{x-3} \right)\left( \text{x+5} \right)}\text{=}\dfrac{\text{1}}{\text{3}}\]

$\Rightarrow 3\left( \text{2x+2} \right)=\left( \text{x-3} \right)\left( \text{x+5} \right)$

$\Rightarrow \text{6x+6=}{{\text{x}}^{\text{2}}}\text{+2x-15}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-4x-21=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-7x+3x-21=0}$

$\Rightarrow \text{x}\left( \text{x-7} \right)\text{+3}\left( \text{x-7} \right)\text{=0}$

$\Rightarrow \left( \text{x-7} \right)\left( \text{x+3} \right)\text{=0}$

$\Rightarrow x\text{=7 or -3}$

Therefore, Rehman’s age is $\text{7 years}$.

 

5. In a class test, the sum of Shefali’s marks in Mathematics and English is $\text{30}$. Had she got $\text{2}$ marks more in Mathematics and $\text{3}$ marks less in English, the product of their marks would have been $\text{210}$. Find her marks in the two subjects.

Ans. Let the marks in maths be $\text{x}$.

Thus, marks in English will be $\text{30-x}$.

Hence, according to question –

$\left( \text{x+2} \right)\left( \text{30-x-3} \right)\text{=210}$

$\left( \text{x+2} \right)\left( \text{27-x} \right)\text{=210}$

$\Rightarrow \text{-}{{\text{x}}^{\text{2}}}\text{+25x+54=210}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-25x+156=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-12x-13x+156=0}$

$\Rightarrow \text{x}\left( \text{x-12} \right)\text{-13}\left( \text{x-12} \right)\text{=0}$

$\Rightarrow \left( \text{x-12} \right)\left( \text{x-13} \right)\text{=0}$

$\Rightarrow \text{x=12,13}$

Case 1- If the marks in mathematics are $\text{12}$ , then marks in English will be $18$.

Case 2- If the marks in mathematics are $\text{13}$ , then marks in English will be $17$.

 

6. The diagonal of a rectangular field is $\text{60}$ metres more than the shorter side. If the longer side is $\text{30}$ metres more than the shorter side, find the sides of the field.

Ans. Let the shorter side of the rectangle be $\text{x m}$.

Thus, Larger side of the rectangle will be $\left( \text{x+30} \right)\text{m}$.

Diagonal of the rectangle be $\sqrt{{{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x+30} \right)}^{\text{2}}}}$

Hence, according to question-

 $\sqrt{{{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x+30} \right)}^{\text{2}}}}\text{=x+60}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x+30} \right)}^{\text{2}}}\text{=}{{\left( \text{x+60} \right)}^{2}}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}\text{+900+60x=}{{\text{x}}^{\text{2}}}\text{+3600+120x}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-60x-2700=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-90x+30-2700=0}$

$\Rightarrow \text{x}\left( \text{x-90} \right)+30\left( \text{x-90} \right)\text{=0}$

$\Rightarrow \left( \text{x-90} \right)\left( \text{x+30} \right)\text{=0}$

$\Rightarrow \text{x=90,-30}$

Since, side cannot be negative. 

Therefore, the length of the shorter side of rectangle is $\text{90 m}$.

Hence, length of the larger side of the rectangle be $\text{120 m}$.

 

7. The difference of squares of two numbers is $\text{180}$. The square of the smaller number is $\text{8}$ times the larger number. Find the two numbers.

Ans. Let the larger number be $\text{x}$ and smaller number be $\text{y}$.

According to question-

${{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=180}$ and ${{\text{y}}^{\text{2}}}\text{=8x}$

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{-8x=180}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{-8x-180=0}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{-18x-10x-180=0}\]

$\Rightarrow \text{x}\left( \text{x-18} \right)+10\left( \text{x-18} \right)\text{=0}$

$\Rightarrow \left( \text{x-18} \right)\left( \text{x+10} \right)\text{=0}$

$\Rightarrow \text{x=18,-10}$

Since, larger cannot be negative as $8$ times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible.

Therefore, the larger number will be $18$.

$\therefore {{\text{y}}^{\text{2}}}\text{=8}\left( \text{18} \right)$

$\Rightarrow {{\text{y}}^{\text{2}}}\text{=144}$

$\Rightarrow \text{y= }\!\!\pm\!\!\text{ 12}$

Hence, smaller number be $\pm 12$.

Therefore, the numbers are $18$ and $12$ or $18$ and $-12$ .

 

8. A train travels $\text{360 km}$km at a uniform speed. If the speed had been $\text{5km/h}$ more, it would have taken $\text{1}$hour less for the same journey. Find the speed of the train.

Ans. Let the speed of the train be $\text{x km/h}$.

Time taken to cover $\text{360 km/h}$ be $\dfrac{\text{360}}{\text{x}}$.

According to question-

$\left( \text{x+5} \right)\left( \dfrac{\text{360}}{\text{x}}\text{-1} \right)\text{=360}$

$\Rightarrow \text{360-x+}\dfrac{\text{1800}}{\text{x}}\text{-5=360}$ 

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{+5x-1800=0}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{+45x-40x-1800=0}\]

$\Rightarrow \text{x}\left( \text{x+45} \right)-40\left( \text{x+45} \right)\text{=0}$

$\Rightarrow \left( \text{x+45} \right)\left( \text{x-40} \right)\text{=0}$

$\Rightarrow \text{x=40,-45}$

Since, the speed cannot be negative.

Therefore, the speed of the train is $\text{40 km/h}$.

 

9. Two water taps together can fill a tank in $\text{9}\dfrac{\text{3}}{\text{8}}$ hours. The tap of larger diameter takes $\text{10}$ 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 the time taken by the smaller pipe to fill the tank be $\text{x hr}$.

So, time taken by larger pipe be $\left( \text{x-10} \right)\text{hr}$.

Part of the tank filled by smaller pipe in $1$ hour is $\dfrac{\text{1}}{\text{x}}$.

Part of the tank filled by larger pipe in $1$ hour is $\dfrac{\text{1}}{\text{x-10}}$.

So, according to question-

$\dfrac{\text{1}}{\text{x}}\text{+}\dfrac{\text{1}}{\text{x-10}}\text{=9}\dfrac{\text{3}}{\text{8}}$

$\Rightarrow \dfrac{\text{1}}{\text{x}}\text{+}\dfrac{\text{1}}{\text{x-10}}\text{=}\dfrac{\text{75}}{\text{8}}$

\[\Rightarrow \dfrac{\text{x-10+x}}{\text{x}\left( \text{x-10} \right)}\text{=}\dfrac{\text{8}}{\text{75}}\]

\[\Rightarrow \dfrac{\text{2x-10}}{\text{x}\left( \text{x-10} \right)}\text{=}\dfrac{\text{8}}{\text{75}}\]

$\Rightarrow \text{75}\left( \text{2x-10} \right)\text{=8}{{\text{x}}^{\text{2}}}\text{-80x}$

$\Rightarrow \text{150x-750=8}{{\text{x}}^{\text{2}}}\text{-80x}$

$\Rightarrow \text{8}{{\text{x}}^{\text{2}}}\text{-230x+750=0}$

$\Rightarrow \text{8}{{\text{x}}^{\text{2}}}\text{-200x-30x+750=0}$

$\Rightarrow \text{8x}\left( \text{x-25} \right)\text{-30}\left( \text{x-25} \right)\text{=0}$

$\Rightarrow \left( \text{x-25} \right)\left( \text{8x-30} \right)\text{=0}$

$\Rightarrow x\text{=25 or }\dfrac{\text{30}}{\text{8}}$

Case 1- If time taken by smaller pipe be $\dfrac{\text{30}}{\text{8}}$ i.e $\text{3}\text{.75 hours}$.

So, Time taken by larger pipe will be negative which is not possible.

Hence, this case is rejected.

Case 2- If the time taken by smaller pipe be $\text{25}$. Then, time taken by larger pipe will be $\text{15 hour}$.

Therefore, time taken by smaller pipe be $\text{25}$ and time taken by larger pipe will be $\text{15 hour}$.

 

10. An express train takes $\text{1}$ hour less than a passenger train to travel $\text{132 km}$ between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speeds of the express train is $\text{11 km/h}$ more than that of the passenger train, find the average speed of the two trains.

Ans. Let the average speed of passenger train be $\text{x km/h}$.

So, Average speed of express train be $\left( \text{x+11} \right)\text{km/h}$.

Thus, according to question.

$\therefore \dfrac{\text{132}}{\text{x}}\text{-}\dfrac{\text{132}}{\text{x+11}}\text{=1}$

$\Rightarrow \text{132}\left[ \dfrac{\text{x+11-x}}{\text{x}\left( \text{x+11} \right)} \right]\text{=1}$

$\Rightarrow \dfrac{\text{132 }\!\!\times\!\!\text{ 11}}{\text{x}\left( \text{x+11} \right)}\text{=1}$

$\Rightarrow \text{132 }\!\!\times\!\!\text{ 11=x}\left( \text{x+11} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+11x-1452=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+44x-33x-1452=0}$

$\Rightarrow \text{x}\left( \text{x+44} \right)\text{-33}\left( \text{x+44} \right)\text{=0}$

$\Rightarrow \left( \text{x+44} \right)\left( \text{x-33} \right)\text{=0}$

$\Rightarrow x\text{=-44 or 33}$

Since, speed cannot be negative.

Therefore, the speed of the passenger train will be $\text{33 km/h}$ and thus, the speed of the express train will be $\text{44 km/h}$.

 

11. Sum of the areas of two squares is $\text{468 }{{\text{m}}^{\text{2}}}$. If the difference of their perimeters are $\text{24 m}$, find the sides of the two squares.

Ans. Let the sides of the two squares be $\text{x m}$ and $\text{y m}$.

Thus, their perimeters will be $\text{4x}$ and $\text{4y}$ and areas will be ${{\text{x}}^{2}}$ and ${{\text{y}}^{2}}$.

Hence, according to question –

$\text{4x-4y=24}$

$\Rightarrow \text{x-y=6}$

$\Rightarrow \text{x=y+6}$

And ${{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}\text{=468}$

Substituting value of x-

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

$\Rightarrow \text{36+}{{\text{y}}^{\text{2}}}\text{+12y+}{{\text{y}}^{\text{2}}}\text{=468}$

$\Rightarrow \text{2}{{\text{y}}^{\text{2}}}\text{+12y-432=0}$

$\Rightarrow {{\text{y}}^{\text{2}}}\text{+6y-216=0}$

$\Rightarrow {{\text{y}}^{\text{2}}}\text{+18y-12y-216=0}$

$\Rightarrow \text{y}\left( \text{y+18} \right)\text{-12}\left( \text{y+18} \right)\text{=0}$

$\Rightarrow \left( \text{y+18} \right)\left( \text{y-12} \right)\text{=0}$

$\Rightarrow \text{y=-18 or 12}$

Since, side cannot be negative.

Therefore, the sides of the square are $\text{12 m}$ and $\left( \text{12+6} \right)\text{m}$ i.e $\text{18 m}$.

 

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.3

Opting for the NCERT solutions for Ex 4.3 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 4.3 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

 

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 10 students who are thorough with all the concepts from the Subject Quadratic Equations textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 10 Maths Chapter 4 Exercise 4.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

 

Besides these NCERT solutions for Class 10 Maths Chapter 4 Exercise 4.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

 

Do not delay any more. Download the NCERT solutions for Class 10 Maths Chapter 4 Exercise 4.3 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

 

NCERT Solutions for Class 10 Maths

 

NCERT Solutions for Class 10 Maths Chapter 4 Exercises

Chapter 4 Quadratic Equations All Exercises in PDF Format

Exercise 4.1

2 Questions and Solutions

Exercise 4.2

6 Questions and Solutions

Exercise 4.4

5 Questions and Solutions

FAQs on NCERT Solutions for Class 10 Maths Chapter 4: Quadratic Equations - Exercise 4.3

1. Give a brief description of the Chapters Quadratic equations.

Ans: In this chapter, students will study the subject and various ways of finding their origin. This chapter also includes some important applications of quadratic equations. In addition to the details about the quadrilateral equation, you will also learn more about the following:-

  • Nature of Roots.

  • The Solutions of Quadratic Equation by Factorisation.

  • The Solutions of Quadratic Equation by Completing the Square.

The chapter takes students on a journey of quadratic equations with examples of everyday life. It helps students understand concepts and relate better. Practising at regular intervals helps students evaluate teaching and develop command on the subject. The chapter soon ends with topics that help the chapter learn at a glance. 

2. Why should I choose Vedantu for preparation?

Ans: NCERT Solutions For Class 10 Maths are prepared in an easy to understand language, and they are crisp and concise. All the questions are accurately answered from the exercise given at the end of the CBSE NCERT Class 10 Maths. Our solutions of NCERT Solutions for Class 10 Maths have been drafted as per the latest CBSE Class 10 Maths Syllabus and CBSE NCERT Class 10 Maths Book.


Our NCERT Solutions gives you a clear idea and understanding of all the important concepts and help you develop a strong conceptual foundation. These solutions cover all possible questions and question types that can be asked in your Class 10 Maths exams.

3. What are the main topics and sub-topics of the Chapter?

Ans: Quadratic Equations is the fourth chapter of CBSE Class 10 Maths syllabus. In this, you shall come across various interesting topics such as introduction to quadratic equations, quadratic polynomial, the standard form of quadratic equation, factorisation method, quadratic equation by the completion of the square method, nature of roots and a quadratic equation in a graphical representation. Our NCERT Solutions for Class 10 Maths Chapter 4 consists of 4 exercises with 24 questions in total. These questions are phrased in an easy and simple language.

4. How many questions are there in Chapter 4 exercise 4.3?

Ans: There are a total of 11 questions in this exercise. In question 1 and 2 you’ll have to find the roots of the given quadratic equation by applying the quadratic formula. Question 1 has 4 sub-questions and question 2 has 4 sub-questions in it. Question 3 has two sub-questions and in both of them, you’ll have to find the roots of the given quadratic equation. 


Questions 4, 5, 8, 9 and 10 are scenario-based questions. You would be given a particular scenario out of which you need to find the value of x. In question 6, 7 and 11 are similar types of questions. These questions are based on the areas of the square in which you have to find the sides of the squares. 

5. What are the important concepts in Chapter 4 of Class 10 Maths?

Ans: Chapter 4 of Class 10 Maths deals with quadratic equations. In this chapter, students will be able to find the roots of quadratic equations by factorisation and learn the definitions and meanings of a quadratic equation. They will also learn how to find quadratic equations by finding the square and nature of roots. To excel in this chapter, download the Class 10 Maths NCERT Solutions Chapter 4 provided by Vedantu (vedantu.com).

6. How many exercises are there in Chapter 4 Class 10 Maths?

Ans: There are four exercises in this chapter. Each one deals with different topics. Exercise 4.3 is based on the nature of roots. Vedantu provides NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations (Ex 4.3) to get easy access to all the questions and answers of Exercise 4.3 from Chapter 4 of Class 10 Maths, prepared by experts to help the students learn and clear all their doubts. These solutions are available free of cost on Vedantu (vedantu.com). You can download it using the Vedantu app as well.

7. Do I need to practice all the questions given in the NCERT Solutions for Exercise 4.3?

Ans: Exercise 4.3 of Chapter 4 in Class 10 Maths NCERT Solutions consists of 11 questions. The NCERT Solutions are prepared by a panel of subject experts. These solutions will help students understand Exercise 4.3 better. Subject experts, keeping students in mind, created these solutions to help students become well-versed in this chapter and secure perfect scores in the exams.

8. How can I understand Exercise 4.3 in Chapter 4 of Class 10 Maths?

Ans: Maths can be difficult if you fail to understand the concepts given in a particular chapter. To solve that issue, Vedantu has created NCERT solutions for all the chapters for students who need guidance. Click on  NCERT Solutions for Class 10 Maths Chapter 4 to download the PDF of the NCERT Solutions for Chapter 4 of Class 10 Maths and become well-versed in all the concepts.

9. What is the best study material for Class 10 Chapter 4 Maths?

Ans: The best study material one can use to prepare for CBSE exams is the NCERT Solutions. As the NCERT Solutions strictly follow the CBSE curriculum, it will be easier for students to understand the concepts. Vedantu provides the best reference material for all the subjects chapter-wise. Visit Vedantu’s website (vedantu.com) to download the PDF of NCERT Solutions for Class 10 Maths Chapter 4 crafted by subject experts for Chapter 4 of Class 10 Maths and ace your exams.