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

Q1 (1) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. $x^2-2x-8$

Answer:

First, factorise the polynomial to know about the zeroes.

We get, x² - 2x - 8 = 0

x² - 4x + 2x - 8 = 0

x(x-4) + 2(x-4) = 0

(x+2)(x-4) = 0

Therefore, the zeroes of the given quadratic polynomial are -2 and 4

$$\begin{gathered} \alpha =-2 \\ ,\beta =4 \end{gathered}$$

VERIFICATION:

Sum of roots:

$\alpha +\beta =-2+4=2$

By formula: $$\begin{gathered} -\frac{coefficientofx}{coefficientof{x}^2} \\ =-\frac{-2}{1} \\ =2 \\ =\alpha +\beta \end{gathered}$$

Hence Verified

Product of roots:

$\alpha \beta =-2\times 4=-8$

By formula: $$\begin{gathered} \frac{constantterm}{coefficientof{x}^2} \\ =\frac{-8}{1} \\ =-8 \\ =\alpha \beta \end{gathered}$$

Hence Verified

Q1 (ii) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. $4s^2-4s+1$

Answer:

First, factorise the polynomial to know about the zeroes.

We get, $4s^2-4s+1=0$

$4s^2-2s-2s+1=0$

$2s(2s-1)-1(2s-1)=0$

$(2s-1)(2s-1)=0$

Therefore, the zeroes of the given quadratic polynomial are 1/2 and 1/2

$$\begin{gathered} \alpha =\frac{1}{2} \\ ,\beta =\frac{1}{2} \end{gathered}$$

VERIFICATION:

Sum of roots:

$\alpha +\beta =\frac{1}{2}+\frac{1}{2}=1$

By formula: $$\begin{gathered} -\frac{coefficientofx}{coefficientof{x}^2} \\ =-\frac{-4}{4} \\ =1 \\ =\alpha +\beta \end{gathered}$$

Hence Verified

Product of roots:

$\alpha \beta =\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

By formula: $$\begin{gathered} \frac{constantterm}{coefficientof{x}^2} \\ =\frac{1}{4} \\ =\alpha \beta \end{gathered}$$

Hence Verified

Q1 (3) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. $6x^2-3-7x$

Answer:

First, factorise the polynomial to know about the zeroes.

We get, 6x² - 3 - 7x = 0

6x² - 7x - 3 = 0

6x² - 9x + 2x - 3 = 0

3x(2x - 3) + 1(2x - 3) = 0

(3x + 1)(2x - 3) = 0

Therefore, the zeroes of the given quadratic polynomial are -1/3 and 3/2

$$\begin{gathered} \alpha =-\frac{1}{3} \\ ,\beta =\frac{3}{2} \end{gathered}$$

VERIFICATION:

Sum of roots:

$\alpha +\beta =-\frac{1}{3}+\frac{3}{2}=\frac{7}{6}$

By formula: $$\begin{gathered} -\frac{coefficientofx}{coefficientof{x}^2} \\ =-\frac{-7}{6} \\ =\frac{7}{6} \\ =\alpha +\beta \end{gathered}$$

Hence Verified

Product of roots:

$\alpha \beta =-\frac{1}{3}\times \frac{3}{2}=-\frac{1}{2}$

By formula: $$\begin{gathered} \frac{constantterm}{coefficientof{x}^2} \\ =\frac{-3}{6} \\ =-\frac{1}{2} \\ =\alpha \beta \end{gathered}$$

Hence Verified

Q1 (4) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. $4u^2+8u$

Answer:

First, factorise the polynomial to know about the zeroes.

We get, 4u ² + 8u = 0

4u(u + 2) = 0

Therefore, the zeroes of the given quadratic polynomial are 0 and -2

$$\begin{gathered} \alpha =0 \\ ,\beta =-2 \end{gathered}$$

VERIFICATION:

Sum of roots:

$\alpha +\beta =0+(-2)=-2$

By formula: $$\begin{gathered} -\frac{coefficientofx}{coefficientof{x}^2} \\ =-\frac{8}{4} \\ =-2 \\ =\alpha +\beta \end{gathered}$$

Hence Verified

Product of roots:

$\alpha \beta =0\times -2=0$

By formula: $$\begin{gathered} \frac{constantterm}{coefficientof{x}^2} \\ =\frac{0}{4} \\ =0 \\ =\alpha \beta \end{gathered}$$

Hence Verified

Q1 (5) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. $t^2-15$

Answer:

First, factorise the polynomial to know about the zeroes.

We get, t² - 15 = 0

$(t-\sqrt{15})(t+\sqrt{15})=0$

Therefore, the zeroes of the given quadratic polynomial are $-\sqrt{15}$ and $\sqrt{15}$

$$\begin{gathered} \alpha =-\sqrt{15} \\ ,\beta =\sqrt{15} \end{gathered}$$

VERIFICATION:

Sum of roots:

$\alpha +\beta =-\sqrt{15}+\sqrt{15}=0$

By formula: $$\begin{gathered} -\frac{coefficientofx}{coefficientof{x}^2} \\ =-\frac{0}{1} \\ =0 \\ =\alpha +\beta \end{gathered}$$

Hence Verified

Product of roots:

$\alpha \beta =-\sqrt{15}\times \sqrt{15}=-15$

By formula: $$\begin{gathered} \frac{constantterm}{coefficientof{x}^2} \\ =\frac{-15}{1} \\ =-15 \\ =\alpha \beta \end{gathered}$$

Hence Verified

Q1 (6) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. $3x^2-x-4$

Answer:

First, factorise the polynomial to know about the zeroes.

We get, 3x ² - x - 4 = 0

3x ² + 3x - 4x - 4 = 0

3x(x + 1) - 4(x + 1) = 0

(3x - 4)(x + 1) = 0

Therefore, the zeroes of the given quadratic polynomial are 4/3 and -1

$$\begin{gathered} \alpha =\frac{4}{3} \\ ,\beta =-1 \end{gathered}$$

VERIFICATION:

Sum of roots:

$\alpha +\beta =\frac{4}{3}+(-1)=\frac{1}{3}$

By formula: $$\begin{gathered} -\frac{coefficientofx}{coefficientof{x}^2} \\ =-\frac{-1}{3} \\ =\frac{1}{3} \\ =\alpha +\beta \end{gathered}$$

Hence Verified

Product of roots:

$\alpha \beta =\frac{4}{3}\times -1=-\frac{4}{3}$

By formula: $$\begin{gathered} \frac{constantterm}{coefficientof{x}^2} \\ =\frac{-4}{3} \\ =\alpha \beta \end{gathered}$$

Hence Verified

Q2 (1) Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. 1/4 , -1

Answer:

Given: $\alpha +\beta =\frac{1}{4}$ and $\alpha \beta$ = -1

The quadratic polynomial equation is ${x}^2-(\alpha +\beta )+\alpha \beta$

Therefore, the polynomial = $x^2-\frac{1}{4}x-1=4x^2-x-4=0$

Thus, the required quadratic polynomial is $4x^2-x-4$

Q2 (2) Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. $\sqrt{2},1/3$

Answer:

Given: $\alpha +\beta =\sqrt{2}$ and $\alpha \beta =\frac{1}{3}$

The quadratic polynomial equation is ${x}^2-(\alpha +\beta )+\alpha \beta$

Therefore, the polynomial = $x^2-\sqrt{2}x+\frac{1}{3}=3x^2-3\sqrt{2}x+1=0$

Thus, the required quadratic polynomial is $3x^2-3\sqrt{2}x+1$

Q2 (3) Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. $0,\sqrt{5}$

Answer:

Given: $\alpha +\beta =0\mathrm{\$}and$\alpha \beta =\sqrt{5}$

The quadratic polynomial equation is ${x}^2-(\alpha +\beta )+\alpha \beta$

Therefore, the polynomial = $x^2-0x+\sqrt{5}=x^2+\sqrt{5}=0$

Thus, the required quadratic polynomial is $x^2+\sqrt{5}$

Q2 (4) Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. 1,1

Answer:

Given: $\alpha +\beta =1$ and $\alpha \beta =1$

The quadratic polynomial equation is ${x}^2-(\alpha +\beta )+\alpha \beta$

Therefore, the polynomial = $x^2-1x+1=x^2-x+1=0$

Thus, the required quadratic polynomial is $x^2-x+1$

Q2 (5) Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. $-\frac{1}{4},\frac{1}{4}$

Answer:

Given: $\alpha +\beta =-\frac{1}{4}\mathrm{\$}and$\alpha \beta =\frac{1}{4}$

The quadratic polynomial equation is ${x}^2-(\alpha +\beta )+\alpha \beta$

Therefore, the polynomial = $x^2-(-\frac{1}{4})x+\frac{1}{4}=4x^2+x+1=0$

Thus, the required quadratic polynomial is $4x^2+x+1$

Q2 (6) Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively. 4,1

Answer:

Given: $\alpha +\beta =4\mathrm{\$}and$\alpha \beta =1$

The quadratic polynomial equation is ${x}^2-(\alpha +\beta )+\alpha \beta$

Therefore, the polynomial = $x^2-4x+1=0$

Thus, the required quadratic polynomial is $x^2-4x+1$