NCERT Solutions for Class 10 Maths Chapter 2: Polynomials - Exercise 2.2

Exercise 2.2 

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\]        

Ans:                  

Given: \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\].

Now factorize the given polynomial to get the roots.

\[\Rightarrow \text{(x}-\text{4)(x+2)}\]

The value of \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\] is zero. 

when \[\text{x}-\text{4=0}\] or \[\text{x+2=0}\]. i.e., \[\text{x = 4}\] or \[\text{x = }-\text{2}\]

Therefore, the zeroes of \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\] are \[\text{4}\] and \[-2\].

Now, Sum of zeroes\[\text{=4}-\text{2= 2 =}-\dfrac{\text{2}}{\text{1}}\text{=}-\dfrac{\text{Coefficient of x}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

 $ \therefore Sum\,of\,zeroes=-\dfrac{Coefficient\,of\,x}{Coefficient\,of\,{{x}^{2}}} $ 

Product of zeroes \[\text{=4 }\!\!\times\!\!\text{ (}-\text{2)=}-\text{8=}\dfrac{\text{(}-\text{8)}}{\text{1}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

 $ \therefore \Pr oduct\,of\,zeroes=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ .

(ii) \[4{{s}^{2}}-4s+1\]

Ans:

Given: \[\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}\]

Now factorize the given polynomial to get the roots.

\[\Rightarrow {{\text{(2s}-\text{1)}}^{\text{2}}}\]

The value of \[\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}\] is zero. 

when \[\text{2s}-\text{1=0}\], \[\text{2s}-\text{1=0}\]. i.e., \[\text{s =}\dfrac{\text{1}}{\text{2}}\] and \[\text{s =}\dfrac{\text{1}}{\text{2}}\]

Therefore, the zeroes of \[\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}\] are \[\dfrac{\text{1}}{\text{2}}\] and \[\dfrac{\text{1}}{\text{2}}\].

Now, Sum of zeroes\[\text{=}\dfrac{\text{1}}{\text{2}}\text{+}\dfrac{\text{1}}{\text{2}}\text{=1=}\dfrac{\text{(}-\text{4)}}{\text{4}}\text{=}\dfrac{-\text{(Coefficient of s)}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}\]

\[\therefore \text{Sum of zeroes =}\dfrac{-\text{(Coefficient of s)}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}\]

Product of zeroes\[=\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}}=\dfrac{\text{1}}{\text{4}}=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}\]

$ \therefore Product of zeroes=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}} $ .

(iii) \[6{{x}^{2}}-3-7x\]

Ans:

Given: \[\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }\]

\[\Rightarrow \text{6}{{\text{x}}^{\text{2}}}-\text{7x}-\text{3}\]

Now factorize the given polynomial to get the roots.

\[\Rightarrow \text{(3x+1)(2x}-\text{3)}\]

The value of  \[\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }\] is zero. 

when \[\text{3x+1=0}\] or \[\text{2x}-\text{3=0}\]. i.e., \[\text{x =}\dfrac{-\text{1}}{\text{3}}\] or \[\text{x =}\dfrac{\text{3}}{\text{2}}\].

Therefore, the zeroes of  \[\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }\] are \[\dfrac{\text{-1}}{\text{3}}\] and \[\dfrac{\text{3}}{\text{2}}\].

Now, Sum of zeroes\[\text{=}\dfrac{-\text{1}}{\text{3}}\text{+}\dfrac{\text{3}}{\text{2}}\text{=}\dfrac{\text{7}}{\text{6}}\text{=}\dfrac{-\text{(}-\text{7)}}{\text{6}}\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

$ \therefore Sum of zeroes\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ 

Product of zeroes\[\text{=}\dfrac{-\text{1}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{2}}\text{=}\dfrac{-\text{3}}{\text{6}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

$ \therefore \text{Product of zeroes=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ 

(iv) \[4{{u}^{2}}+8u\]                          

Ans:

Given: \[\text{4}{{\text{u}}^{\text{2}}}\text{+8u}\]

\[\Rightarrow \text{4}{{\text{u}}^{\text{2}}}\text{+8u+0}\] 

\[\Rightarrow \text{4u(u+2)}\]

The value of \[\text{4}{{\text{u}}^{\text{2}}}\text{+8u}\] is zero. 

when \[\text{4u=0}\] or \[\text{u+2=0}\]. i.e., \[\text{u = 0}\] or \[\text{u =}-\text{2}\]

Therefore, the zeroes of \[\text{4}{{\text{u}}^{\text{2}}}\text{+8u}\] are \[\text{0}\] and \[\text{-2}\].

Now, Sum of zeroes\[\text{=0+(}-\text{2)=}-\text{2=}\dfrac{-\text{8}}{\text{4}}\text{=}\dfrac{-\text{(Coefficient of u)}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}\]

$ \therefore \text{Sum of zeroes=}\dfrac{-\text{(Coefficient of u)}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}} $ 

Product of zeroes\[\text{=0 }\!\!\times\!\!\text{ (}-\text{2)= 0 =}\dfrac{\text{0}}{\text{4}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}\]

$ \therefore \text{Product of zeroes=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}} $ 

(v) \[{{t}^{2}}-15\]

Ans:

Given: \[{{\text{t}}^{\text{2}}}-\text{15}\]

\[\Rightarrow {{\text{t}}^{\text{2}}}-\text{0t}-\text{15}\] 

Now factorize the given polynomial to get the roots.

\[\Rightarrow \text{(t}-\sqrt{\text{15}}\text{)(t +}\sqrt{\text{15}}\text{)}\] 

The value of \[{{\text{t}}^{\text{2}}}-\text{15}\] is zero. 

when \[\text{t}-\sqrt{\text{15}}\text{=0}\] or \[\text{t+}\sqrt{\text{15}}\text{=0}\], i.e., \[\text{t=}\sqrt{\text{15}}\] or \[\text{t=}-\sqrt{\text{15}}\]

Therefore, the zeroes of \[{{\text{t}}^{\text{2}}}-\text{15}\] are \[\sqrt{\text{15}}\] and \[-\sqrt{\text{15}}\].

Now, Sum of zeroes\[\text{=}\sqrt{\text{15}}\text{+(}-\sqrt{\text{15}}\text{)=0=}\dfrac{-\text{0}}{\text{1}}\text{=}\dfrac{-\text{(Coefficient of t)}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}\]

$ \therefore \text{Sum of zeroes =}\dfrac{-\text{(Coefficient of t)}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}} $ 

Product of zeroes\[\text{=}\left( \sqrt{\text{15}} \right)\times \left( -\sqrt{\text{15}} \right)\text{=}-\text{15=}\dfrac{-\text{15}}{\text{1}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}\]

$ \therefore \text{Product of zeroes =}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}} $ .

(vi) \[3{{x}^{2}}-x-4\]

Ans: 

Given: \[\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\]

Now factorize the given polynomial to get the roots.

 $ \Rightarrow \left( 3x-4 \right)(x+1) $ 

The value of \[\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\] is zero.

when \[\text{3x}-\text{4=0}\] or \[\text{x+1=0}\], i.e., \[\text{x=}\dfrac{\text{4}}{\text{3}}\] or \[\text{x=}-\text{1}\]

Therefore, the zeroes of \[\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\] are \[\dfrac{\text{4}}{\text{3}}\] and \[\text{-1}\].

Now, Sum of zeroes\[\text{=}\dfrac{\text{4}}{\text{3}}\text{+(}-\text{1)=}\dfrac{\text{1}}{\text{3}}\text{=}\dfrac{-\text{(}-\text{1)}}{\text{3}}\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

$ \therefore \text{Sum of zeroes=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ 

Product of zeroes\[\text{=}\dfrac{\text{4}}{\text{3}}\times \text{(}-\text{1)=}\dfrac{-\text{4}}{\text{3}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

$ \therefore \text{Product of zeroes =}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ .

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. 

(i) \[\dfrac{1}{4},-1\]  

Ans:

Given: \[\dfrac{\text{1}}{\text{4}}\text{,-1}\]

Let the zeroes of polynomial be \[\text{ }\!\!\alpha\!\!\text{ }\] and \[\text{ }\!\!\beta\!\!\text{ }\].

Then, 

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{  =}\dfrac{\text{1}}{\text{4}}\]

\[\text{ }\!\!\alpha\!\!\text{  }\!\!\beta\!\!\text{ =}-\text{1}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

$ \Rightarrow {{x}^{2}}-\dfrac{1}{4}x-1 $ 

$ \Rightarrow 4{{x}^{2}}-x-4 $ 

Therefore, the quadratic polynomial is \[\text{4}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\].

(ii) \[\sqrt{2},\dfrac{1}{3}\]      

Ans:

Given: \[\sqrt{\text{2}}\text{,}\dfrac{\text{1}}{\text{3}}\]

Let the zeroes of polynomial be \[\text{ }\!\!\alpha\!\!\text{ }\] and \[\text{ }\!\!\beta\!\!\text{ }\].

Then,

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{  =}\sqrt{\text{2}}\]

\[\text{ }\!\!\alpha\!\!\text{  }\!\!\beta\!\!\text{ =}\dfrac{\text{1}}{\text{3}}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

\[\Rightarrow {{\text{x}}^{\text{2}}}-\sqrt{2}\text{x+}\dfrac{1}{3}\]

\[\Rightarrow 3{{\text{x}}^{\text{2}}}-3\sqrt{2}\text{x+1}\]

Therefore, the quadratic polynomial is \[3{{\text{x}}^{\text{2}}}-3\sqrt{2}\text{x+1}\].

(iii) \[0,\sqrt{5}\]  

(here, root is missing)  

Ans:

Given: \[\text{0,}\sqrt{\text{5}}\]

Let the zeroes of polynomial be \[\text{ }\!\!\alpha\!\!\text{ }\] and \[\text{ }\!\!\beta\!\!\text{ }\].

Then,

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{  = 0}\]

\[\text{ }\!\!\alpha\!\!\text{  }\!\!\beta\!\!\text{ =}\sqrt{\text{5}}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

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

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\sqrt{5}\]

Therefore, the quadratic polynomial is \[{{\text{x}}^{\text{2}}}\text{+}\sqrt{\text{5}}\].

(iv) \[1,1\]     

Ans:

Given: \[\text{1,1}\]

Let the zeroes of polynomial be \[\text{ }\!\!\alpha\!\!\text{ }\] and \[\text{ }\!\!\beta\!\!\text{ }\].

Then,

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{  =1}\]

\[\text{ }\!\!\alpha\!\!\text{  }\!\!\beta\!\!\text{  =1}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

\[\Rightarrow {{\text{x}}^{\text{2}}}-\text{1x+1}\]

Therefore, the quadratic polynomial is \[{{\text{x}}^{\text{2}}}-\text{x+1}\].

(v) \[-\dfrac{1}{4},\dfrac{1}{4}\]      

Ans:

Given: \[-\dfrac{\text{1}}{\text{4}}\text{,}\dfrac{\text{1}}{\text{4}}\]

Let the zeroes of polynomial be \[\text{ }\!\!\alpha\!\!\text{ }\] and \[\text{ }\!\!\beta\!\!\text{ }\].

Then,

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ =}-\dfrac{\text{1}}{\text{4}}\]

\[\text{ }\!\!\alpha\!\!\text{  }\!\!\beta\!\!\text{  =}\dfrac{\text{1}}{\text{4}}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

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

 $ \Rightarrow \text{4}{{\text{x}}^{\text{2}}}\text{+ x +1} $ 

Therefore, the quadratic polynomial is  $ \text{4}{{\text{x}}^{\text{2}}}\text{+ x +1} $ .

(vi) \[4,1\]

Ans: 

Given: \[\text{4,1}\]

Let the zeroes of polynomial be \[\text{ }\!\!\alpha\!\!\text{ }\] and \[\text{ }\!\!\beta\!\!\text{ }\].

Then,

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{  = 4}\]

\[\text{ }\!\!\alpha\!\!\text{  }\!\!\beta\!\!\text{  =1}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

$ \Rightarrow {{\text{x}}^{\text{2}}}-\text{4x+1} $ 

Therefore, the quadratic polynomial is  $ {{\text{x}}^{\text{2}}}-\text{4x+1} $.

Importance of Polynomials

Polynomials are expressions having more than two algebraic terms. They can also be defined as the sum of several terms where the same variable/variables has/have different powers. 

They are important because they serve as the language in most Mathematical expressions. They are used to represent appropriate relations between different variables or numbers. We encourage students to learn from this chapter to be able to solve tricky problems easily in exams.

NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.2

Definition 

If x be a variable and x be a positive integer with a1, a2,  a3, ……., an be constants then, f(x) = anxn + an-1x-1 + ….. + a1x + a0   is called a polynomial in x.

In this polynomial anxn + an-1x-1, ………, a1x and a0 are called the terms of the polynomial, and an, an-1,…., a1, a0 are called their coefficients.

Example: p(x) = 4x -1  is a polynomial in variable x.

         q(y) = 2y2 – 3y + 5 is a polynomial in variable y. 

Note: Expressions like 3x2 - 2√x + 7, 1/(3x-5), etc. are not polynomials as some powers of x are not integers. 

Degree of a Polynomial

The exponent of the highest degree term in a polynomial is called its degree. 

Example: 

1. f(x) = 4x -1 is a polynomial in x of degree 1. 

2. q(y) = 2y2 - 3y + 5 is a polynomial in y of degree 2. 

Types of Polynomial

Let us see the different types of polynomials.

1. Constant Polynomial: It is a polynomial of degree zero.

Ex: f(x) = 5, g(x) = 2, etc.

The constant polynomial f(x) = 0 is called the zero polynomial. The degree of the zero polynomial is not defined as f(x) = 0, g(x) =0x, h(x) = 0x2, p(x) = 0x3, etc. are all equal to the zero polynomial. 

2. Linear Polynomial: It is a polynomial of degrees.

Ex: f(x) = 4x -1.

A polynomial q(y) = 3y2 + 4 is not linear as it has degree 2.

In general, any linear polynomial in x with real coefficients is of the form f(x) = ax + b where a & b are real numbers and a 0. 

Note: A linear polynomial may be a monomial (having only one term. Ex: q(x) = 5x) or a binomial 

(having 2 terms. Ex: f(x) = 4x-1)

3. Quadratic Polynomial: It is a polynomial of degree 2.

Ex: f(x) = 5x2 - 3x + 4.

In general, a quadratic polynomial is of the form f(x) = ax2 + bx + c, where a, b & c are real numbers and a 0. 

4. Cubic Polynomial: A polynomial of degree 3 is called a cubic polynomial. 

Ex: f(x) = 2x3 – 5x2 + 7x – 9.

The general form of a cubic polynomial is f(x) = ax3 + bx2 + cx + d, where a, b, c, d are real numbers and a 0.  

Value of a Polynomial

If f(x) is a polynomial and ‘a’ is a real number then the real number obtained by replacing x by ‘a’ , is called the value f(x) at x = a and is denoted by f(a). For example: if f(x) = 2x2 - 3x -2, then its value 

1. at x = 1 is given by f(1) = 2(1)2 - 3 1 -2 = 2-3-2 = -3

2. at x = -2, is given by f(-2) = 2(-2)2 -3(-2)-2 = 8 + 6 -2 = 12

Zero of a Polynomial

Consider the cubic polynomial f(x) = x3 – 6x2 + 11x - 6. The value of this polynomial at x =1, 2, 3 are respectively 

f(1) =13 – 6 * 12 + 11 * 1 – 6 = 1-6 + 11 -6 =0

f(2) = 23 – 6 * 22 + 11 * 2 – 6 = 8 – 24 + 22 – 6 = 0

f(3) = 33 – 6* 32 + 11 * 3 -6 = 27  -54 + 33 -6 = 0

1, 2, 3 are called the zeroes of the cubic polynomial f(x) = x3 -6x2 + 11x - 6.

A real number is 0 of a polynomial f(x), if f() = 0.

To find this we solve the equation f(x) = 0.

For example: if the polynomial is f(x) = x -3 then putting x -3 = 0, we get x = 3

Then the real number 3 is the zero of the polynomial f(x) = x -3.

There can be more than one zero of a polynomial as can be seen in the example f(x) = x3-6x2 + 11x - 6.

Graph of a Polynomial

The graph of a polynomial f(x) is the set of all points (x, y), where y = f(x). This graph is a smooth free hand curve, passing through the points (x1, y1), (x2, y2), (x3,y3), ……, etc. y1, y2, y3, …… are the estimated values of the polynomial f(x) at x1, x2, x3,..... respectively. 

Geometrical meaning of the zeroes of a linear polynomial

We take an example f(x) = 4x - 2

Let y  4x - 2

We make a table with values of y corresponding to different values of x. 

(Image to be added soon)

The points A(0, 1) and B (2, 6) are plotted on the graph paper on a suitable scale. A line drawn runs through these points to derive the graph of the given polynomial. It can be seen that the graph intersects the x-axis at C (½, 0), where y = f(x) = 0 so that the zero of the given polynomial is ½. 

Meaning of the Zeroes of a Quadratic Polynomial

Graph of a quadratic polynomial f(x) = x2 - 2x - 8

Let y = x2 - 2x - 8

We construct a table containing the values of y corresponding to various values of x:

(Image to be added soon)

We now plot the points (-4, 16), ( -3, 7), (-2, 0) , (-1, -5), (0 -8), (1, 9), (2, 8), (3, -5), (4, 0) , (5, 7) and (6, 16) on a graph paper and draw a smooth free hand curve passing through these points. Hence, the curve derived demonstrates the graph of the polynomial f(x) = x2 -2x -8. Such a curve is called a parabola. The lowest point of the graph (1, 9) is called the vertex of the parabola. 

The Following Observations Can Be Made

1. The graph is symmetrical about the line parallel to the y-axis through the vertex (1, -9). This line is called the axis of the parabola.

2. The concavity of the parabola is towards the positive direction of the y-axis (this happens when the coefficient of x2 is positive.

3. As  x2 -2x -8 = (x-4) (x + 2), the polynomial f(x) has two factors (x-4) & (x+2), so that the parabola cuts the x-axis at two distinct points (4,0) & (-2, 0). The x-coordinates of these two points: 4 & -2 are the two zeroes of f(x).

Note: 

1. The number of zeroes of a polynomial depends on its degree. For a polynomial of degree n, the number of zeroes can be utmost n or less. For example, for a quadratic, polynomial the no. of zeroes can be 2 or less.

2. For a quadratic polynomial, its graph (ex: a parabola)  may cut the x-axis at two points in fig (i), or at one point in fig (ii) or at no point at all in fig (iii).

(Image to be added soon)

Solved Examples

Q1. If 2 and -2 are zeroes of polynomial Ax4 + bx3 + Cx2 + Dx + E, then show that 16A + 4C + E = 8B + 2D = 0.

Solution: 

p(x) = Ax4 + bx3 + Cx2 + Dx + E

As, p(2) = 0

A(2)4 + B(2)3 +C(2)2 + D(2) + E = 0

16A + 8B + 4C + 2D + E = 0

Also, p(-2) = 0

16A - 8B + 4C - 2D + E = 0

16A + 4C + E = 8B + 2D

From (i), we get 

(16A + 4C + E) + 8B + 2D = 0

(8B + 2D) + 8B + 2D = 0

From (ii)

From (ii)

2(8B + 2D = 0) …… (iii)

Also, 16A + 4C + E = 8B + 2D          

From (ii)

From (ii)

16A + 4C + E = 0 …… (iv)

From (iii) and (iv), we get

16A + 4C + E = 8B + 2D = 0

Q2. Sum of two zeroes of a polynomial of degree 4 is -1 and their product is -2. If the other two zeroes are √5 and √-5. Find the polynomial.

Solution:

Two zeroes of polynomial p(x) are √5 and √-5.

∴ (x - √5) and (x + √5) are factors of p(x).

∴ (x - √5)  (x + √5) is a factor of p(x).

∴ x2 - 5 is a factor pf p(x).

A polynomial with the sum of zeroes -1 and product -2 is given by x2 + x -2.

∴ x2 + x -2 is also factor of p(x).

Since, degree of p(x) is 4

∴ p(x) = (x2 - 5 )(x2 + x -2)

= x4 + x3 - 7x2 - 5x + 10.

Q3. Figure shows the graph of the polynomial f(x) = ax2 + bx + c for which 

(Image to be added soon)

(a) a < 0, b> 0

(b) a < 0, b < 0

(c) a > 0, b < 0

(d) a > 0, b > 0

Solution: 

(b) : y = f(x) = ax2 + bx + c represents a parabola opening downward.

∴ a < 0

Also, point P(-b/2a,  (4ac - b2)/4a) lies in the second quadrant. 

∴ - b/2a < 0 b < 0

∵ a<0

NCERT Solutions Class 10 Maths All Chapters

• Chapter 1 - Real Numbers

• Chapter 2 - Polynomials

• Chapter 3 - Pair of Linear Equations in Two Variables

• Chapter 4 - Quadratic Equations

• Chapter 5 - Arithmetic Progressions

• Chapter 6 - Triangles

• Chapter 7 - Coordinate Geometry

• Chapter 8 - Introduction to Trigonometry

• Chapter 9 - Some Applications of Trigonometry

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Areas Related to Circles

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

NCERT Solutions Class 10 Maths Chapter 2 Exercises