## Book: RD Sharma - Mathematics

### Chapter: 2. Polynomials

#### Subject: Maths - Class 10th

##### Q. No. 1 of Exercise 2.1

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##### Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:(i) (ii) (iii) (iv) (v) (vi)(vii) (viii)

(i)

⇒ x2 - 4x + 2x – 8

⇒ x (x - 4) + 2 (x - 4)

For zeros of f(x), f(x) = 0

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

x = -2, 4

Therefore zeros of the polynomial are -2 & 4

Sum of zeros = -2 + 4 = 2 = -(-2) = 2 =

Product of zeros = -2 × 4 = -8 = -8 =

(ii)

⇒ 4s2 -2s - 2s + 1

⇒ 2s (2s - 1) -1 (2s - 1)

For zeros of g(s), g(s) = 0

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

s =

Therefore zeros of the polynomial are ,

Sum of zeros = + = 1 = -(- )= 1 =

Product of zeros = × = = =

(iii)

For zeros of h(t), h(t) = 0

t2 = 15

t = ± √15

Therefore zeros of t = √15 & -√15

Sum of zeros = √15 + (- √15) = 0 = 0 =

Product of zeros = × = -15 = -15 =

(iv) f(x) = x

⇒ 6x2 - 7x -3

⇒ 6x2 - 9x + 2x - 3

⇒ 3x(2x - 3) +1(2x - 3)

⇒ (3x + 1) (2x - 3)

For zeros of f(x), f(x) = 0

⇒ (3x + 1) (2x - 3) = 0

x =

Therefore zeros of the polynomial are

Sum of zeros = = = = =

Product of zeros = × = = =

(v) P (x) = x2 + 3√2x - √2x - 6

For zeros of p(x), p(x) = 0

⇒ x (x + 3√2) -√2 (x + 3√2) = 0

⇒ (x - √2) (x + 3√2) = 0

x = √2, -3√2

Therefore zeros of the polynomial are √2 & -3√2

Sum of zeros = √2 -3√2 = -2√2 = -2√2 =

Product of zeros = √2 × -3√2 = -6 = -6 =

(vi) q (x) = √3x2 + 10x + 7√3

⇒ √3x2 + 10x + 7√3

⇒ √3x2 + 7x + 3x + 7√3

⇒ √3x (x +) + 3 (x + )

⇒ (√3x + 3) (x +)

For zeros of Q(x), Q(x) = 0

(√3x + 3) (x +) = 0

X = ,

Therefore zeros of the polynomial are ,

Sum of zeros = + =

Product of zeros = = × = 7 =

(vii) f(x) = x2 - (√3 + 1)x + √3

f(x) = x2 - √3x - x + √3

f(x) = x(x - √3) -1(x - √3)

f(x) = (x - 1) (x - √3)

For zeros of f(x), f(x) = 0

(x - 1) (x - √3) = 0

X = 1, √3

Therefore zeros of the polynomial are 1 & √3

Sum of zeros =

Product of zeros = 1 × √3 = √3 = √3 =

(viii) g(x) = a(x2 + 1) – x(a2 + 1)

g(x) = ax2 - a2x – x + a

g(x) = ax(x - a) -1(x - a)

g(x) = (ax - 1) (x - a)

For zeros of g(x), g(x) = 0

(ax - 1) (x - a) = 0

X = , a

Therefore zeros of the polynomial are & a

Sum of zeros =

Product of zeros = × a = 1 = 1 =

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