HBSE 10th Class Maths Solutions Chapter 2 Polynomials Ex 2.2

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) x² – 2x – 8
(ii) 4s² – 4s + 1
(iii) 6x² – 3 – 7x
(iv) 4u² + 8u
(v) t² – 15
(vi) 3x² – x – 4.
Solution :
(i) We have, f(x) = x² – 2x – 8
= x² – (4 – 2)x – 8
= x² – 4x + 2x – 8
= (x² – 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) = 4s² – 4s + 1
= 4s² – (2 + 2)s + 1
= 4s² – 2s – 2s + 1
= (4s² – 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) = 6x² – 3 – 7x
= 6x² – 7x – 3
= 6x² – (9 – 2) x – 3
= 6x² – 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) = 4u² + 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) = t² – 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) = 3x² – x – 4
= 3x² – (4 – 3)x – 4
= 3x² – 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) = x² – (α + β)x + α × β
= x² – \(\frac{1}{4}\) x – 1
= \(\frac{1}{4}\) (4x² – x – 4)
∴ Quadratic polynomial = 4x² – 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) = x² – (α + β)x + α × β
= x² – √2x + \(\frac{1}{3}\)
= \(\frac{1}{3}\) (3x² – 3√2x + 1)
∴ Quadratic polynomial = 3x² – 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) = x² – (α + β)x + α × β
= x² – 0 × x + √5
= x² +√5

(iv) We have,
Sum of zeroes = α + β = 1
Product of zeroes = α × β = 1
So, the required quadratic polynomial is f(x) = x² – (α + β)x + α × β
= x² – 1.x + 1
= x² – 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) = x² – (α + β)x + α × β
= x² – (- \(\frac{1}{4}\)) x + \(\frac{1}{4}\)
= x² + \(\frac{1}{4}\) x + \(\frac{1}{4}\)
= \(\frac{1}{4}\) (4x² + x + 1)
∴ Quadratic polynomial = 4x² + x + 1.

(vi) We have,
Sum of zeroes = α + β = 4
Product of zeroes = α × β = 1
So, the required polynomial is
f(x) = x² – (α + β)x + α × β
= x² – 4x + 1