RD Sharma Solutions Chapter 2 Polynomials Exercise 2.1 Class 10 Maths
Chapter Name | RD Sharma Chapter 2 Polynomials |
Book Name | RD Sharma Mathematics for Class 10 |
Other Exercises |
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Related Study | NCERT Solutions for Class 10 Maths |
Exercise 2.1 Solutions
1. Find the zeroes of each of the following quadratic polynomials and verify the relationship between the zeroes and their co efficient :
(i) f(x) = x2 – 2x – 8
(ii) g(s) = 4s2 – 4x + 1
(iii) h(t) = t2 – 15
(iv) p(x) = x2 + 2√2x + 6
(v) q(x) = √3x2 + 10x + 7√3
(vi) f(x) = x2 – (√3 + 1)x + √3
(vii) g(x) = a(x2 + 1) – x(a2 + 1)
(viii) 6x2 – 3 – 7x
Solution
(i) f(x) = x2 – 2x – 8
f(x) = x2 – 2x – 8 = x2 – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x + 2) (x – 4)
Zeroes of the polynomials are –2 and 4
Sum of the zeroes = –(co efficient of x)/(co efficient of x)
– 2+ 4 = – (–2)/1
⇒ 2 = 2
Product of the zeroes = (constant term)/(co efficient of x2 )
= 24 = –8/1
⇒ –8 = –8
∴ Hence the relationship verified
(ii) 9(5) = 45 - 45 + 1 = 452 - 25 - 25 + 1 = 25(25 - 1) - 1(25 - 1)
= (25 - 1)(25 - 1 )
Zeroes of the polynomials are 1/2 and 1/2
Sum of zeroes = (–co efficient of s)/(co efficient of s2 )
1/2 + 1/2 = -(-4)/4
⇒ 1 = 1
Product of the zeroes = (constant term)/(co efficient of s2 )
1/2 ×1/2 = 1/4
⇒ 1/4 = 1/4
∴ Hence the relationship verified.
(iii) h(t) = t2 - 15 = (t2) - (√15)2 = (t + √15)(t - √15)
zeroes of the polynomials are -√15 and √15
sum of zeroes = 0
- √15 + √15 = 0
⇒ 0 = 0
Product of zeroes = -15/1
-√15 × √15 = - 15
⇒ -15 = -15
∴ Hence the relationship verified.
(iv) p(x) = x2 + 2√2x - 6 = x2 + 3√2x + √2 × 3√2
= x(x + 3√2) - √2( 2 + 3√2) = (x - √2)(x + 3√2)
Zeroes of the polynomial are 3√2 and -3√2
Sum of the zeroes = -3√2/1
√2 - 3√2 = -2√2
⇒ -2√2 = -2√2
Product of zeroes = √2
⇒ √2 × -3√2 = -6/1
-6 = -6
Hence, the relationship verified.
(v) 2(x) = √3x2 + 10x + 7√3 = √3x2 + 7x + 3x + 7√3
= √3x(x + √3) + 7(x +√3)
= (√3x + 7)(x + √3)
Zeroes of the polynomials are - √3, -7/√3
Sum of zeroes = -10/√3
⇒ -√3 - 7/√3 = -10/√3
Product of zeroes = 7√3/3
⇒ (√3x - 7)/√30 = 7
⇒ 7 = 7
Hence, relationship verified.
(vi) f(x) = x2 - (√3 + 1)x + √3 = x2 - √3x - x + √3
= x(x - √3) - 1(x - √3)
= (x - 1)(x - √3)
Zeroes of the polynomials are 1 and √3
Sum of zeroes = -( co efficient of x)/(co efficient x2) = -(-√3-1)/1
1 + √3 = √3 +1
Product of zeroes = (constant term)/(co efficient of x2) = √3/1
1 × √3 = √3 = √3 = √3
∴ Hence, relationship verified.
(vii) g(x) = a[(x2 + 1) - x(a2 + 1)]2 = ax2 + a - a2 x - x
= ax2 - [(a2 + 1) -x] + 0 = ax2 - a2 x - x + a
= ax(x -a) - 1(x-a) = (x - a)(ax - 1)
Zeroes of the polynomials = 1/a and a
Sum of the zeroes = -[-a2 - 1]/a
⇒ 1/a + a = (a2 + 1)/a
⇒ (a2 + 1)/a = (a2 + 1)/a
Product of zeroes = a/a
⇒ 1/a × a = a/a
⇒ (a2 + 1)/a = (a2 + 1)/a
Product of zeroes = a/a
⇒ 1 = 1
Hence, relationship verified
(viii) 6x2 - 3 - 7x
= 6x2 - 7x - 3
= (3x + 11)(2x - 3)
Zeroes of polynomials are +3/2 and -1/3
Sum of zeroes = -1/3 + 3/2
= 7/6
= -(-7)/6
= -(co efficeint of x)/(co efficient of x2 )
Product of zeroes = -1/3 × 3/2
= -1/2
= -3/6
= (constant term)/(co efficient of x2 )
∴ Hence, relationship verified.
2. If α and β are the zero of the quadratic polynomial f(x) = ax2 + bx + c, then evaluate:
(i) α - β
(ii) 1/α - 1/ß
(iii) 1/α + 1/ß - 2αß
(iv) α2 ß+ αß2
(v) α4 + ß4
(vi) 1/(aα + b) + 1/(aß + b)
(vii) ß/(aα+b) + α/(aß+b)
(viii) a[α2/ß + ß2/α] + b[α/a + ß/a]
Solution
f(x) = ax2 + bx + c
α + ß = -b/a
αß = c/a
since α + ß are the roots (or) zeroes of the given polynomials
(i) α - ß
The two zeroes of the polynomials are 
(iii) 1/α + 1/ß - 2αß
⇒ [(α+ß)/αß] - 2αß
⇒ -b/a × a/c - 2c/a
= -2c/a - b/c
= (-ab - 2c2)/ac - [b/c + 2c/a]
(iv) a2ß + αß2
αß(α + ß)
= c/a(-b/a)
= -bc/a2 
Since α and ß are the zeroes of the given polynomial
∴ Sum of zeroes [α + ß] = -1/6
Product of zeroes (αß) = - 1/3
= α/ß + ß/α = (α2 + ß2 )/αß = [(α+ß)2 - 2αß]/αß
Since α + ß are the zeroes of the polynomial : x2 - x - 4
Sum of the roots (α + ß) = 1
Product of the roots(αß) = -4
1/α + 1/ß - αß = (α+ß)/αß - αß
= 1/-4 + 4
= -1/4 + 4
= (-1 + 16)/4
= 15/4
Since α and ß are the roots of the polynomial : 4x2 - 5x - 1
∴ Sum of the roots α + ß = 5/4
Product of the roots αß = -1/4
Hence α2ß + αß2 = αß(α + ß) = 5/4(-1/4) = -5/16
Since α and ß are the roots of the polynomial x2 + x - 2
Product of roots αß = -2
Since α and ß are the roots of the quadratic polynomial
f(x)= x2 - 5x + 4
Sum of roots = α + ß = 5
Product of roots = αß = 4
1/α + 1/ß - 2αß
= (ß+α)/αß - 2αß
= 5/4 - 2× 4
= 5/4 - 8
= -27/4
Since α and ß are the zeroes of the polynomial f(t) = t2 - 4t + 3
Since α + ß = 4
Product of zeroes αß = 3
Hence α4ß3 + α3ß4 = α3ß3 (α + ß) = [3]2 [4] = 108
Since α and ß are the zeroes of the polynomials
p(y) = 5y2 - 7y + 1
Sum of the zeroes αß = 1/6
Product of zeroes αß = 1/6
1/α + 1/ß
= (α+ß)/αß
= (7×5)/(5×1)
= 7
Since α and ß are the zeroes of the polynomials
Sum of the zeroes α + ß = 6/3
Product of the zeroes αß = 4/3 
α2/ß2 + ß2 + α2 = p2/q2 - 4p2/q + 2
Since α and ß are the roots of the polynomials
f(x) = x2 - px + 2
sum of zeroes = p = α + ß
Product of zeroes = q = αß
Let the two zeroes of the polynomial be α and ß
f(x) = x2 + px + 45
sum of the zeroes = -p
Product of zeroes = 45
⇒ (α - ß)2 - 4αß = 144
⇒ p2 - 4×45 = 144
⇒ p2 = 324
p = ±1
Let the two zeroes of the f(t) = kt2 + 2t + 3k be α and ß
Sum of the zeroes (α + ß)
Product of the zeroes αß
-2/k = 3k/k
⇒ -2k = 3k2
⇒ k(3k + 2) = 0
⇒ k = 0
⇒k = -2/3
Let the two zeroes of one polynomial.
⇒ α × Î± = -9/4
⇒ tα2 = +9/4
⇒ α = +3/2
Sum of zeroes = 8k/4 = 0
Hence 8k = 0
or k = 0
f(x) = x2 - 1
Sum of zeroes α + ß = 0
Product of zeroes αß = -1
Sum of zeroes = 2α/ß+ 2ß/α = (2α2 + 2ß2)/αß 
Hence the quadratic equation is x2 - (sum of zeroes)x + product of zeroes
= k(x2 + 4x + 14)
f(x) = x2 - 3x - 2
Sum of zeroes [α +ß] = 3
Product of zeroes [αß] = -2
Sum of zeroes = 1/[2α + ß] + 1/[2ß+α]
α +ß = 24
αß = 8
.............
2α = 32
α = 16
ß = 8
αß = 16 × 8 = 128
Quadratic equation
⇒ x2 - (sum of zeroes) + product of zeroes
⇒ k[x2 - 24x + 128]
f(x) = x2 - p(x+1)c = x - px = -p -c
Sum of zeroes = α + ß = p
Product of zeroes = - p- c = αß
(α + 1+ß) = αß + α +ß + 1 = -p - c +p + 1
= 1 - c = R.H.S
∴ Hence, proved.
(i) α + 2, ß+2
(ii) (α - 1)/(α+1), (ß - 1)/(ß+1)
f(x) = x2 - 2x + 3
Sum of zeroes = 2 = (α + ß)
Product of zeroes = 3 = (αß)
(i) sum of zeroes = (α + 2)+(ß+2) = α + ß+4 = 2+ 4 = 6
Product of zeroes = (α +2)(ß+2)
= αß+2α+2ß+4 = 3+2(2) + 4 = 11
Quadratic equation = x2 - 6x + 11 = k[x2 - 6x + 11]
f(x) = x2 + p +q
Sum of zeroes = p = α + ß
Sum of the new polynomial = (α + ß)2 + (α - ß)2
= (-p)2 + α2 + ß2 - 2αß
= p2 + (α + ß)2 - 2αß - 2αß
= p2 + p2 - 4q
= 2p2 - 4q
Product of zeroes = (α+ß)2 × (α - ß)2
= [-p]2 × (p2 - 4q) = (p2 - 4q)p2
Quadratic equation = x2 - [2p2 - 4q]+ p2 [-4q+ p]
f(x) = k{x2 - 2(p2 - 28)x + p2 (q2 - 4q)}