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

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

VERIFICATION:

Sum of roots:

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

By formula: $\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-2}{1}\\ =2\\=\alpha +\beta$

Hence Verified

Product of roots:

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

By formula: $\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-8}{1}\\ =-8\\=\alpha \beta$

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

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

VERIFICATION:

Sum of roots:

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

By formula: $\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-4}{4}\\ =1\\=\alpha +\beta$

Hence Verified

Product of roots:

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

By formula: $\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{1}{4}\\ =\alpha \beta$

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

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

VERIFICATION:

Sum of roots:

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

By formula: $\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-7}{6}\\ =\frac{7}{6}\\=\alpha +\beta$

Hence Verified

Product of roots:

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

By formula: $\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-3}{6}\\ =-\frac{1}{2}\\=\alpha \beta$

Hence Verified

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

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

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

VERIFICATION:

Sum of roots:

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

By formula: $\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{8}{4}\\ =-2\\=\alpha +\beta$

Hence Verified

Product of roots:

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

By formula: $\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{0}{4}\\=0\\ =\alpha \beta$

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 $\sqrt{15}$

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

VERIFICATION:

Sum of roots:

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

By formula: $\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{0}{1}\\ =0\\=\alpha +\beta$

Hence Verified

Product of roots:

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

By formula: $\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-15}{1}\\=-15\\ =\alpha \beta$

Hence Verified

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

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

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

VERIFICATION:

Sum of roots:

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

By formula: $\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-1}{3}\\ =\frac{1}{3}\\=\alpha +\beta$

Hence Verified

Product of roots:

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

By formula: $\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-4}{3}\\ =\alpha \beta$

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\$ 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}\$ and $\alpha \beta =\frac{1}{4}$

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

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

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