Chapter 2 - Polynomials, RD Sharma Solutions - (Part-1) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 2.33

Ques.1. Find the zeros of each of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:
(i) f(x) = x² − 2x − 8
(ii) g(s) = 4s² − 4s + 1
(iii) h(t) = t² − 15
(iv) 6x² − 3 − 7x
(v) p(x)= x²+ 2√2x − 6  
(vi) q(x)= √3x² +10x +7√3
(vii) f(x)= x² −(√3 + 1) x + √3
(viii) g(x) = a(x² + 1) − x(a² + 1)
(ix) h(s)= 2s² − (1+2√2)s + √2
(x) f(v) = v² + 4√3v −15
(xi) p(y) = y² + (3√5/2)y - 5
(xii) q(y) = 7y² − (11/3)y - 2/3
Ans.1. 
(i) We have,
f(x) = x² − 2x − 8
f(x) = x² + 2x − 4x − 8
f(x) = x (x + 2) − 4(x + 2)
f(x) = (x + 2) (x − 4)
The zeros of f(x) are given by
f(x) = 0
x² − 2x − 8 = 0
(x + 2) (x − 4) = 0
x + 2 = 0
x = −2
Or
x − 4 = 0
x = 4
Thus, the zeros of f(x) = x² − 2x − 8 are α = −2 and β = 4
Now,
Sum of the zeros = α + β
= (-2) + 4
= -2 + 4
= 2
and
= (-Coefficient of x/Coefficient of x²)
= -(-2/1)
= 2
Therefore, sum of the zeros = -(Coefficient of x/Coefficient of x²)
Product of the zeros = αβ
= -2 x 4
= -8
and
= Constant term/Coefficient of x²
= -8/1
= -8
Therefore,
Product of the zeros = Constant term/Coefficient of x^{2
Hence, the relationship between the zeros and coefficient are verified.}
(ii) Given g(s) = 4s² - 4s + 1
When have,
g(s) = 4s² − 4s + 1
g(s) = 4s² − 2s − 2s + 1
g(s) = 2s (2s − 1) − 1(2s − 1)
g(s) = (2s − 1) (2s − 1)
The zeros of g(s) are given by
g(s) = 0
4s² - 4s + 1 = 0
(2s - 1)(2s - 1) = 0
(2s - 1) = 0
2s = +1

Or
(2s - 1) = 0
2s = 1
s = (1/2)
Thus, the zeros of g(x) = 4s² - 4s + 1 are
α = 1/2 and β = 1/2
Now, sum of the zeros = α + β
= (1/2 + 1/2)
= (1+1)/2
= 2/2
= 1
and
= (-Coefficient of x/Coefficient of x²)
= -((-4)/(4))
= 4/4
= 1
Therefore, sum of the zeros = (-Coefficient of x/Coefficient of x²)
Product of the zeros = αβ
= (1/2 x 1/2)
= 1/4
and = Constant term/Coefficient of x² 
= 1/4
Therefore, the product of the zeroes = Constant term/Coefficient of x² 
Hence, the relation-ship between the zeros and coefficient are verified.

(iii) Given h(t) = t² - 15
We have,
h(t) = t² − 15
h(t) = (t)² − (√15)²
h(t) = (t + √15) (t −√15)
The zeros of h(t) are given by
h(t) = 0
(t − √15) (t + √15) = 0
(t − √15) = 0
t = √15
or
(t + √15) = 0
t = −√15
Hence, the zeros of h(t) are α = √15 and  β = −√15
Now,
Sum of the zeros = α + β
= √15 + (-√15)
= √15 - √15
= 0
and = (-Coefficient of x/Coefficient of x²)
= 0/1
= 0
Therefore, sum of the zeros = (-Coefficient of x/Coefficient of x²)
also,
Product of the zeros = αβ
= √15 x -√15
= -15
and,
Constant term/Coefficient of x² 
= -15/1
= -15
Therefore, the product of the zeros = Constant term/Coefficient of x² 
Hence, The relationship between the zeros and coefficient are verified.

(iv) Given f(x) = 6x² - 3 - 7x
We have, f(x) = 6x² - 7x - 3
f(x) = 6x² - 9x + 2x - 3
f(x) = 3x(2x - 3) + 1(2x - 3)
f(x) = (3x + 1)(2x - 3)
The zeros of f(x) are given by
f(x) = 0
6x² - 7x - 3 = 0
(3x + 1)(2x - 3) = 0
3x + 1 = 0
3x = -1
x = -1/3
Or
2x - 3 = 0
2x = 3
x = 3/2
Thus, the zeros of  f(x) = 6x² - 7x - 3 are α = -1/3 and β = 3/2.
Now,
Sum of the zeros = α + β
= (-1/3 + 3/2)

= (-2/6 + 9/6)
=(-2 + 9)/(6)
= 7/6
and, =  (-Coefficient of x/Coefficient of x²)
Product of the zeros = α × β

= (-1)/2
and,= Constant term/Coefficient of x² 
= (-3)/6
= (-1)/2
Product of zeros = Constant term/Coefficient of x²
Hence, the relation between the zeros and its coefficient are verified.

(v) Given p(x) = x² + 2√2x - 6
We have,
p(x) = x² + 2√2x - 6
p(x) = x² + 3√2x - √2x - 6
p(x) = x(x + 3√2)-√2(x + 3√2)
p(x) = (x - √2)(x + 3√2)
The zeros of p(x) are given by
p(x) = 0
p(x) = x² + 2√2x - 6
x² + 2√2x - 6 = 0
(x - √2)(x + 3√2) = 0
(x - √2) = 0
x = √2
Or
(x + 3√2) = 0
x = -3√2
Thus, The zeros of p(x) = x² + 2√2 - 6 are α = √2 and β = -3√2
Now,
Sum of the zeros = α + β
= √2 - 3√2
= +√2(1 - 3)
= √2(-2)
= -2√2
and, (-Coefficient of x/Coefficient of x²)
= ((-2√2)/(1))
= -2√2
Therefore, Sum of the zeros = (-Coefficient of x/Coefficient of x²)
Product of the zeros = α x β
= √2 x -3√2
= -3 x 2
= -6
and
Constant term/Coefficient of x² 
= -6/1
= -6
Therefore, The product of the zeros = Constant term/Coefficient of x² 
Hence, the relation-ship between the zeros and coefficient are verified.

(vi) Given q(x) = √3x² + 10x + 7√3
We have, q(x) = √3x² + 10x + 7√3

The zeros of g(x) are given by
g(x) = 0
√3x² + 10x + 7√3 = 0
(x + √3)(√3x + 7) = 0
x + √3= 0
x = -√3
Or
√3 + 7 = 0
√3x = -7
x = (-7)/(√3)
Thus, the zeros of q(x) = √3x² + 10x + 7√3 are α = -√3 and β = -7/√3
Now,
Sum of the zeros = α + β


= (-10)/(√3)
and = (-Coefficient of x/Coefficient of x²)

Therefore, sum of the zeros = (-Coefficient of x/Coefficient of x²)
Product of zeros = α × β

= +7
and = -(Coefficient of x/Coefficient of x²)

= 7
Therefore, the product of the zeros = Constant term/Coefficient of x² 
Hence, the relationship between the zeros and coefficient are verified.

(vii) Given f(x) = x² - (√3 + 1) + √3
f(x) = x² - √3x - 1x  + √3
f(x) = x(x -√3) -1(x - √3)
f(x) = (x - 1)(x - √3)
The zeros of ƒ(x) are given by
f(x) = 0
x² - (√3 + 1)x + √3 = 0
(x - 1)(x - √3) = 0
(x - 1) = 0
x = 0 + 1
x = 1
Or
x - √3 = 0
x = 0 + √3
x = √3
Thus, the zeros of x² - (√3 + 1) are α = 1 and β = √3
Now,
Sum of zeros = α + β
= 1 + √3
= 1 + √3
And,
= (-Coefficient of x/Coefficient of x²)

Therefore, sum of the zeros = (-Coefficient of x/Coefficient of x²)
Product of the zeros = αβ
= 1 x √3
= √3
And
= Constant term/Coefficient of x²
= √3/1
= √3
Product of zeros = Constant term/Coefficient of x²
Hence, the relation-ship between the zeros and coefficient are verified.

(viii) Given g(x) = a(x² + 1) -x(a² + 1)
g(x) = ax² -xa² + a - x
g(x) = xa (x - a) -1 (x - a)
g(x) = (xa - 1)(x - a)
The zeros of g(x) are given by
g(x) = 0
ax² - (a² + 1)x + a = 0
xa - 1 =0
xa = 1
x = 1/a
or
x - a = 0
x = a
Thus, the zeros of g(x) = ax² -(a² + 1) x + a are
α = 1/a and  β = a
Sum of the zeros = α + β
= (1/a) + a

and = (-Coefficient of x/Coefficient of x²)

Product of the zeros = α x β

= 1
And, = Constant term/Coefficient of x²
= a/a

= 1
Therefore,
Product of the zeros = Constant term/Coefficient of x²
Hence, the relationship between the zeros and coefficient are verified.

(ix) h(s) = 2s² − (1 + 2√2)s + √2
h(s) = 2s² − s − 2√2s + √2
h(s) = s(2s − 1) − √2(2s − 1)
h(s) = (2s − 1)(s − √2)
The zeros of h(s) are given by
h(s) = 0
2s² − (1 + 2√2) s + √2 = 0
(2s − 1)(s − √2) = 0
(2s − 1) = 0 or (s − √2) = 0
s = 1/2 or s = √2
Thus, the zeros of h (s) = (2s - 1)(s - √2) are α = 1/2 and β = √2
Now,
Sum of the zeros = α+β
= 1/2 + √2
and
(-Coefficient of s/Coefficient of s²)


Therefore, sum of the zeros = (-Coefficient of x/Coefficient of x²)
Product of the zeros = αβ

and
Constant term/Coefficient of s²

Therefore,
Product of the zeros = Constant term/Coefficient of x²
Hence, the relation-ship between the zeros and coefficient are verified.

(x) f(v) = v² + 4√3v −15
f(v) = v² + 5√3v − √3v − 15
= v² − √3v + 5√3v − 15
= v(v − √3) + 5√3 (v − √3)
= (v - √3)(v + 5√3)
The zeros of f(v) are given by
f(v) = 0
v² + 4√3v − 15 = 0
(v + 5√3)(v − √3) = 0
(v − √3) = 0 or (v + 5√3) = 0
v = √3 or v = −5√3
Thus, the zeros of f(v) = (v − √3)(v + 5√3) are α = √3 and β = −5√3.
Now,
Sum of the zeros = α + β
= √3 −5√3= −4√3
and


Therefore, sum of the zeros = (-Coefficient of x/Coefficient of x²)
Product of the zeros  = αβ
=√3 × (−5√3) = −15
and
Constant term / Coefficient of v²
= -15/1 = -15
Therefore,
Product of the zeros = Constant term/Coefficient of s²
Hence, the relationship between the zeros and coefficient are verified.

(xi) h(s) = 2s² − (1 + 2√ 2)s + √2
h(s) = 2s² − s − 2√2s + √2
h(s) =s  (2s − 1) − √2(2s − 1)
h(s) = (2s −1)(s − √2)
The zeros of h(s) are given by
h(s) = 0
2s² − (1 + 2√2)s +√2 = 0
(2s − 1)(s − √2) = 0
(2s − 1) = 0 or (s − √2) = 0

Thus, the zeros of h(s) = (2s − 1)(s − √2) are α = 1/2 and β = √2
Now,
Sum of the zeros = α + β

and
−Coefficient of s/Coefficient of s²

Therefore, sum of the zeros = −Coefficient of x/Coefficient of x²
Product of the zeros = αβ

and
Constant term/Coefficient of s²
Therefore,
Product of the zeros = Constant term/Coefficient of s²
Hence, the relationship between the zeros and coefficient are verified.

(x) f(v) = v² + 4√3v − 15
f(v) = v² + 5√3v − √3v − 15
= v² − √3v + 5√3v − 15
= v(v − √3) + 5√3 (v − √3)
= (v − √3)(v + 5√3)
The zeros of f(v) are given by
f(v) = 0
v² + 4√3v − 15 = 0
(v + 5√3)(v − √3) = 0
(v − √3) = 0 or (v + 5√3) = 0
v = √3 or v = −5√3
Thus, the zeros of f(v) = (v − √3)(v + 5√3) are α = √3 and β = −5√3.
Now,
Sum of the zeros = α + β
=√3 − 5√3 = −4√3
and
−Coefficient of v/Coefficient of v²

Therefore, sum of the zeros = −Coefficient of x/Coefficient of x²
Product of the zeros = αβ
= √3 × (−5√3) = −15
and
Constant term/Coefficient of v²
= -15/1 = −15
Therefore,
Product of the zeros = Constant term/Coefficient of x²
Hence, the relation-ship between the zeros and coefficient are verified.

(xi) 
p(y)

The zeros are given by p(y) = 0.
Thus, the zeros of p(y)= 1/2 (2y − √5)(y + 2√5) are α
Now,
Sum of the zeros = α + β

and
−Coefficient of y/Coefficient of y²

Therefore, sum of the zeros = −Coefficient of x/Coefficient of x²
Product of the zeros = αβ

= −5
and
Constant term/Coefficient of y²
= -5/1 = -5
Therefore,
Product of the zeros = Constant term/Coefficient of x²
Hence, the relationship between the zeros and coefficient are verified.

(xii) 

= 1/3 (21y² − 14y + 3y − 2)
= 1/3 [7y(3y − 2) + 1 (3y − 2)]
= 1/3 [(7y + 1)(3y − 2)]
The zeros are given by q(y) = 0.
Thus, the zeros of q(y) = 1/3 (7y + 1)(3y − 2) are α =  -1/7 and β = 2/3.
Now,
Sum of the zeros = α + β

= 11/21
and
−Coefficient of y/Coefficient of y²

Therefore, sum of the zeros = −Coefficient of x/Coefficient of x²
Product of the zeros = αβ

= -2/21
and
Constant term/Coefficient of y²

= -2/21
Therefore,
Product of the zeros = Constant term/Coefficient of x²
Hence, the relationship between the zeros and coefficient are verified.