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

  • Exercise 2.2
  • Exercise 2.3

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 



3. If α and ß are the zeros of the quadratic polynomial f(x) = 6x2 + x - 2, find the value of α/ß + ß/α 
Solution
f(x)  = 6x2 - x - 2 
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

4. If α and ß are the zeros of the quadratic polynomial f(x) = x2 - x + 4, find the value of 1/α + 1/ß - αß. 
Solution

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 


5. If α and ß  are the zeros of the quadratic polynomial p(x) = 4x2 - 5x - 1, find the value of α2 ß + αß2.
Solution

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 


6. If α and ß are the zeros of the quadratic polynomial f(x) = x2 + x - 2, find the value of 1/α - 1/ß. 
Solution

Since α and ß are the roots of the polynomial x2  + x - 2 

∴ Sum of roots α + ß = 1 
Product of roots αß = -2
⇒ -1/ß

7. If α and ß are the zeros of the quadratic polynomial f(x) = x2 - 5x + 4, find the value of 1/α - 1/ß - 2αß.
Solution

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 


8. If α and ß  are the zeros of the quadratic polynomial f(t) = t2  - 4t + 3, find the value of α4ß3 + α3ß4 .
Solution

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 


9. If α and ß are the zeros of the quadratic polynomial p(y) = 5y2 - 7y + 1, find the value of 1/α + 1/ß.
Solution

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


10. If α and ß are the zeros of the quadratic polynomial p(s) = 3s2 - 6s + 4, find the value of α+ß + ß/α + 2[1/α + 1/ß] + 3αß
Solution

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 


11. If α and ß are the zeros of the quadratic polynomial f(x) = x2 - px + q, prove that 
α22 + ß2 + α2 = p2/q2 - 4p2/q + 2
Solution

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 


12. If the squared difference of the zeros of the quadratic polynomial f(x) = x2 + px + 45 is equal to 144. Find the value of p. 
Solution

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 = 144 + 180 
⇒ p2 = 324 
p = ±1

13. If the sum of the zeroes of the quadratic polynomial f(t) = kt2 + 2t + 3k is equal to their product, find the value of k. 
Solution

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 

⇒ 2k + 3k2  = 0 
⇒ k(3k + 2) = 0 
⇒ k = 0 
⇒k = -2/3 

14. If one zero of the quadratic polynomial f(x) = 4x2 - 8kx - 9 is negative of the other, find the value of k. 
Solution

Let the two zeroes of one polynomial. 

f(x) = 4x2 - 5k - 9 be α, -α
⇒ α × α = -9/4
⇒ tα2  = +9/4
⇒ α = +3/2
Sum of zeroes = 8k/4 = 0 
Hence 8k = 0 
or k = 0 

15. If α and ß are the zeroes of the quadratic polynomial f(x) = x2 - 1, find a quadratic polynomial whose zeroes are 2α/ß and 2ß/α.
Solution

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) 


16. If α and ß are the zeroes of the quadratic polynomial f(x) = x2 - 3x - 2, find a quadratic polynomial whose zeroes are 1/(2α + ß) + 1/(2ß+α). 
Solution

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]


17. If α and ß are the zeros of a quadratic polynomial such that α + ß  = 24 and α - ß = 8, find a quadratic polynomial having α and ß as its zeros. 
Solution

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]


18. If α and ß are the zeroes of the quadratic polynomial f(x) = x2 - p(x+ 1) - c, show that (α +1)(ß + 1) = 1- c.
Solution

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.


19. If α and ß are the zeros of the quadratic polynomial f(x) = x2 - 2x +3, find a polynomial whose roots are : 
(i) α + 2, ß+2 
(ii) (α - 1)/(α+1), (ß - 1)/(ß+1)
Solution

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]

(ii) sum of zeroes = (α-1)/(α+1) + (ß-1)/(ß+1) 

20. If α and ß are the zeroes of the polynomial f(x) = x2 + px + q, form a polynomial whose zeroes are (α + ß)2 and (α - ß)2 . 
Solution

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)}

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