Haryana State Board HBSE 10th Class Maths Solutions Chapter 2 Polynomials Ex 2.2 Textbook Exercise Questions and Answers.

## Haryana Board 10th Class Maths Solutions Chapter 2 Polynomials Exercise 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) We have, f(x) = x^{2} – 2x – 8

= x^{2} – (4 – 2)x – 8

= x^{2} – 4x + 2x – 8

= (x^{2} – 4x) + (2x – 8)

= x(x – 4) + 2(x – 4)

= (x – 4) (x + 2)

To find the zeroes of the polynomial f(x), put f(x) = 0

⇒ (x – 4) (x + 2) = 0

⇒ x = 4, – 2

Therefore, the zeroes of polynomial f(x) are : α = 4, β = – 2.

Now, Sum of the zeroes = α + β = (4 – 2)

= 2 = \(\frac{-(-2)}{1}\)

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

Product of zeroes = α × β

= 4 × (- 2) = \(-\frac{8}{1}\)

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

Hence, zeroes of the polynomial f(x) are 4, – 2.

(ii) We have,

f(s) = 4s^{2} – 4s + 1

= 4s^{2} – (2 + 2)s + 1

= 4s^{2} – 2s – 2s + 1

= (4s^{2} – 2s) – (2s – 1)

= 2s (2s – 1) – 1 (2s – 1)

= (2s – 1) (2s – 1)

To find the zeroes of the polynomial f(s), put f(s) = 0.

⇒ (2s – 1) (2s – 1) = 0

⇒ s = \(\frac{1}{2}\), \(\frac{1}{2}\)

Therefore, zeroes of the polynomial f(s) are α = \(\frac{1}{2}\), β = \(\frac{1}{2}\).

∴ It has equal roots.

Now, Sum of zeroes = α + β

= \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1

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

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

Product of zeroes = α × β

= \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\)

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

Hence, zeroes of polynomial f(s) = \(\frac{1}{2}\), \(\frac{1}{2}\).

(iii) We have,

p(x) = 6x^{2} – 3 – 7x

= 6x^{2} – 7x – 3

= 6x^{2} – (9 – 2) x – 3

= 6x^{2} – 9x + 2x – 3

= 3x (2x – 3) + 1 (2x – 3)

= (2x – 3) (3x + 1)

To find the zeroes of the polynomial p(x)

put p(x) = 0

⇒ (2x – 3) (3x + 1) = 0

⇒ x = \(\frac{3}{2}\), – \(\frac{1}{3}\)

Therefore, zeroes of the polynomial p(x) are : α = \(\frac{3}{2}\), β = – \(\frac{1}{3}\)

Now, Sum of zeroes = α + β

= \(\frac{3}{2}\) – \(\frac{1}{3}\)

= \(\frac{9-2}{6}=\frac{7}{6}=\frac{-(-7)}{6}\)

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

Product of zeroes = α × β

= \(\frac{3}{2} \times\left(-\frac{1}{3}\right)=-\frac{3}{6}\)

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

Hence, zeroes of the polynomial p(x) = \(\frac{3}{2}\), – \(\frac{1}{3}\)

(iv) We have,

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

= 4 u(u + 2)

To find the zeroes of the polynomial f(u), put f(u) = 0

⇒ 4 u(u + 2) = 0

⇒ u = 0, – 2.

Therefore, zeroes of the polynomial f(u) are : α = 0, and β = – 2.

Now, Sum of zeroes = α + β

= 0 + (- 2) = – 2 = \(\frac{-8}{4}\)

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

Product of zeroes = α × β

= 0 × (- 2)

= \(\frac{0 \times(-8)}{4}=\frac{0}{4}\)

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

Hence, zeroes of the polynomial f(u) = 0, – 2.

(v) We have,

f(t) = t^{2} – 15

= (t)^{2} – (√15)^{2}

= (t + √15) (t – √15)

To find the zeroes of polynomial f(t), put f(t) = 0.

⇒ (t + √15) (t – √15) = 0.

⇒ t = – √15, √15

Therefore, zeroes of polynomial f(t) are :

α = – √15 and β = √15

Now, Sum of zeroes = α + β

= – √15 + √15 = 0 = – \(\frac{0}{1}\)

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

Product of zeroes = α × β

= – √15 × √15

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

Hence, zeroes of the polynomial f(t) = – √15 , √15

(vi) We have,

f(x) = 3x^{2} – x – 4

= 3x^{2} – (4 – 3)x – 4

= 3x^{2} – 4x + 3x – 4

= x(3x – 4) + 1 (3x – 4)

= (3x – 4) (x + 1)

To find the zeroes of the polynomial f(x),

put f(x) = 0

⇒ (3x – 4) (x + 1) = 0

⇒ x = \(\frac{4}{3}\), – 1

Therefore, zeroes of the polynomial f(x) are : α = \(\frac{4}{3}\) and β= – 1

Now, Sum of zeroes = α + β

= \(\frac{4}{3}\) – 1

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

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

Product of zeroes = α × β

= \(\frac{4}{3}\) × (- 1)

= – \(\frac{4}{3}\)

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

Hence, zeroes of polynomial f(x) = \(\frac{4}{3}\) , – 1.

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.

Solution :

(i) We have,

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

Product of zeroes = α × β = – 1

So, the required quadratic polynomial is f(x) = x^{2} – (α + β)x + α × β

= x^{2} – \(\frac{1}{4}\) x – 1

= \(\frac{1}{4}\) (4x^{2} – x – 4)

∴ Quadratic polynomial = 4x^{2} – x – 4.

(ii) We have,

Sum of zeroes = (α + β) = √2

Product of zeroes = α × β = \(\frac{1}{3}\)

So, the required quadratic polynomial is :

p(x) = x^{2} – (α + β)x + α × β

= x^{2} – √2x + \(\frac{1}{3}\)

= \(\frac{1}{3}\) (3x^{2} – 3√2x + 1)

∴ Quadratic polynomial = 3x^{2} – 3√2x + 1

(iii) We have,

Sum of zeroes = (α + β) = 0.

Product of zeroes = α × β = √5

So, the required quadratic polynomial is

f(x) = x^{2} – (α + β)x + α × β

= x^{2} – 0 × x + √5

= x^{2} +√5

(iv) We have,

Sum of zeroes = α + β = 1

Product of zeroes = α × β = 1

So, the required quadratic polynomial is f(x) = x^{2} – (α + β)x + α × β

= x^{2} – 1.x + 1

= x^{2} – x + 1

(v) We have,

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

Product of zeroes = α × β = \(\frac{1}{4}\)

So, the required quadratic polynomial is f(x) = x^{2} – (α + β)x + α × β

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

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

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

∴ Quadratic polynomial = 4x^{2} + x + 1.

(vi) We have,

Sum of zeroes = α + β = 4

Product of zeroes = α × β = 1

So, the required polynomial is

f(x) = x^{2} – (α + β)x + α × β

= x^{2} – 4x + 1