Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 2 Polynomials Ex 2.2 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2

Question 1.

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

(i) x^{2} – 2x – 8

(ii) 4s^{2} – 4s + 1

(iii) 6x^{2} – 3 – 7x

(iv) 4u^{2} + 8u

(v) t^{2} – 15

(vi) 3x^{2} – x -4

Solution:

(i) Given Quadratic polynomial,

x^{2} – 2x – 8

∵ [S = -2, P = -8]

= x^{2} – 4x + 2x – 8

= x (x – 4) + 2 (x – 4)

= (x – 4) (x + 2)

The value of x^{2} – 2x – 8 is zero,

iff (x – 4) = 0 or (x + 2) = 0

iff x = 4 or x = – 2

Therefore, zeroes of x2 – 2x – 8 are – 2 and 4.

Now, Sum of zeroes = (- 2) + (4) = 2

= \(\frac{-(-2)}{1}=-\frac{(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}}\)

Product of zeroes = (- 2) (4) = – 8

= \(\frac{-8}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)

Hence, relationship between the zeroes and the coefficient are verified.

(ii) Given quadratic polynomial,

4s^{2} – 4s + 1

= 4s^{2} – 2s – 2s + 1

∵ [S = -4, P = 4 × 1]

= 2s (2s – 1) – 1 (2s – 1)

= (2s – 1) (2s – 1)

The value of 4s^{2} – 4s + 1 is zero

iff (2s – 1) = 0 or (2s – 1) = 0

iff s = \(\frac{1}{2}\) or s = \(\frac{1}{2}\)

Therefore, zeroes of 4s^{2} – 4s + 1 are \(\frac{1}{2}\) and \(\frac{1}{2}\)

Now, sum of zeroes = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1

= \(\frac{-(-4)}{4}=\frac{-(\text { Coefficient of } s)}{\left(\text { Coefficient of } s^{2}\right)}\)

Product of Zeroes = \(\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4}\)

= \(\frac{\text { Constant term }}{\text { Coefficient of } s^{2}}\)

Hence, relationship between the zeroes and the coefficients are verified.

(iii) Given quadratic polynomial,

6x^{2} -3 – 7x

= 6x^{2} – 7x – 3

∵ [S = – 7, P = 6x-3=-18]

= 6x^{2} -9x + 2x-3

= 3x (2x – 3) + 1 (2x – 3)

= (2x – 3) (3x + 1)

The value of 6x^{2} – 3 – 7x is zero

iff (2x – 3) = 0 or 3x + 1 = 0

iff x = \(\frac{3}{2}\) or x = –\(\frac{1}{3}\)

Therefore, zeroes of 6x^{2} – 3 – 7x are \(\frac{3}{2}\) and –\(\frac{1}{3}\)

Now, Sum of zeroes = \(\frac{1}{3}\)

= \(\frac{3}{2}+\left(\frac{-1}{3}\right)\)

= \(\frac{3}{2}-\frac{1}{3}=\frac{9-2}{6}\)

= \(\frac{7}{6}=\frac{-(-7)}{6}\)

= \(\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}}\)

Product of zeroes = \(\left(\frac{3}{2}\right)\left(\frac{-1}{3}\right)\)

= \(\frac{-3}{6}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)

Hence, relationship between the zeroes and the coefficients are verified.

(iv) Given quadratic polynomial,

4u^{2} + 8u = 4u(u + 2)

The value of 4u^{2} + 8 u is zero

iff 4u = 0 or u + 2 = 0

iff u = 0 or u = – 2

Therefore, zeroes of 4u^{2} + 8M are 0 and – 2

Now, Sum of zeroes = 0 + (- 2) = -2

= \(\frac{-8}{4}\)

= \(-\frac{(\text { Coefficient of } u)}{\text { Coefficient of } u^{2}}\)

Product of zeroes = (0) (- 2) = 0

= \(\frac{0}{4}=\frac{\text { Constant term }}{\text { Coefficient of } u^{2}}\)

Hence, relationship between the zeroes and the coefficients are verified.

(v) Given quadratic polynomial,

t^{2} – 15 = t^{2} – (√15)^{2}

= (t – √15) (t + √15)

The value of t^{2} – 15 is zero

iff t – √15 = 0 or t + √15 = 0

iff t = √15 or t = – √15

Therefore, zeroes of t^{2} – 15 are – √15 and √15.

Now, Sum of zeroes = -√15 + √15 = 0 = \(\frac{0}{1}\)

= \(\frac{-(\text { Coefficient of } t)}{\text { Coefficient of } t^{2}}\)

Product of zeroes = (-√15) (√15) = – 15 = \(-\frac{15}{1}\)

= \(\frac{0}{1}\)

= \(=\frac{\text { Constant term }}{\text { Coefficient of } t^{2}}\)

Hence, relationship between the zeroes and the coefficients are verified.

(vi) Given quadratic polynomial,

3x^{2} – x – 4

= 3x^{2} + 3x – 4x – 4

= 3x (x + 1) – 4 (x + 1)

∵ [S = – 1, P = 3 x – 4 = – 12]

= (x + 1) (3x – 4)

The value of 3x^{2} – x – 4 is zero

iff (x + 1) = 0 or 3x – 4 = 0

iff x = -1 or x = \(\frac{4}{3}\)

Therefore, zeroes of 3x^{2} – x – 4 are – 1 and \(\frac{4}{3}\)

Now, Sum of zeroes = – 1 + \(\frac{4}{3}\)

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

= \(\frac{-(-1)}{3}=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\)

Product of zeroes = (- 1) \(\frac{4}{3}\)

= \(\frac{-4}{3}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)

Hence, relationship between the zeroes and the coefficients are verified.

Question 2.

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

(i) \(\frac{1}{4}\), -1

(ii) √2, \(\frac{1}{3}\)

(iii) 0, √5

(iv) 1, 1

(v) –\(\frac{1}{4}\), \(\frac{1}{4}\)

(vi) 4, 1. [MQP 2015]

Solution:

(i) Given that, sum of zeroes and products of zeroes of given polynomial \(\frac{1}{4}\) are -1 respectively.

Let the quadratic polynomial be ax^{2} + bx + c and its zeroes be α and β

α + β = Sum of zeroes = \(\frac{1}{4}\)

and αβ = Product of zeroes = – 1

Now, ax^{2} + bx + c = k (x – α) (x – β) where k is any constant

= k [x^{2} – (α + β)x + αβ]

= k[x^{2} – \(\frac{1}{4}\)x + (-1)]

= k[x^{2} – \(\frac{1}{4}\)x – 1]

for different value of k, we get different quadratic polynomials.

(ii) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are √2 and \(\frac{1}{3}\) respectively.

Let the quadratic polynomial be ax^{2} + bx + c and its zeroes be α and β

α + β = Sum of zeroes = √2

and αβ = Product of zeroes = \(\frac{1}{3}\)

Now, ax^{2} + bx + c = k (x – α) (x – β) where k is any constant

= k[x^{2} – (α + β) x + αβ]

= k[x^{2} – √2x + \(\frac{1}{3}\)]of k, we get different quadratic polynomial.

for different values of k, we get different quadratic polynomial.

(iii) Given that, sum of zeroes and products of zeroes of given quadratic polynomial are 0 and √5 respectively.

Let the quadratic polynomial be ax^{2} + bx + c and its zeroes be α and β

α + β = Sum of zeroes = 0

and αβ = Product of zeroes = √5

Now, ax^{2} + bx + c = k(x – α) (x – β)

where k is any constant

= k [x^{2} – (a + (α+ β)x + αβ)

= k[x^{2} – 0x + √5]

= k[x^{2} + √5]

for different values of k, we get different quadratic polynomial.

(iv) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are 1 and 1 respectively.

Let the quadratic polynomial be ax^{2} + bx + c and its zeroes be α and β

α + β = Sum of zeroes = 1 and

αβ = Product of zeroes = 1

Now, ax^{2} + bx + c = k(x – α) (x – β)

where k is any constant.

= k [x^{2} – (α + β)x + αβ]

= k [x^{2} – 0x + √5]

= it [x^{2} – x + √5]

for different values of k, we get different quadratic polynomial.

(v) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are \(-\frac{1}{4}\) and \(\frac{1}{4}\) respectively.

Let the quadratic polynomial be ax^{2} + bx + c and its zeroes be α and β

α + β = Sum of zeroes = –\(\frac{1}{4}\)

and αβ = Product of zeroes = \(\frac{1}{4}\)

Now ax^{2} + bx + c = k (x – α) (x – β) where k is any constant

= k[x^{2} – (α + β) x + αβ]

= k[x^{2} – (-\(\frac{1}{4}\))x + \(\frac{1}{4}\)]

= k[x^{2} + \(\frac{1}{4}\)x + \(\frac{1}{4}\)]

For different values of k, we get different quadratic polynomial.

(vi) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are 4 and 1 respectively. Let the quadratic polynomial be

ax^{2} + bx + c and its zeroes be α and β

α + β = sum of zeroes = 4 and

αβ = Product of zeroes = 1

Now, ax2 + bx + c = k(x – α) (x – β) where k is any constant

= k [x^{2} – (α + β) x + αβ]

= k [x^{2} – 4x + 1]

For different values of k, we get different quadratic polynomials.