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Class 10 RD Sharma Solutions- Chapter 2 Polynomials – Exercise 2.2
  • Difficulty Level : Medium
  • Last Updated : 13 Jan, 2021

Q1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeros and coefficients in each of the following cases:

(i) f(x) = 2x3 + x2 – 5x + 2; 1/2, 1, -2

Solution:

For the followings numbers to be the zeros of polynomial they must satisfy the following equation i.e. f(x)=0 so now checking:

When x = 1/2

 f(1/2) = 2(1/2)3 + (1/2)2 – 5(1/2) + 2 

f(1/2) = 1/4 + 1/4 – 5/2 + 2 = 0 



f(1/2) = 0,    

Hence, x = 1/2 is a zero of the given polynomial.

When x = 1

f(1) = 2(1)3 + (1)2 – 5(1) + 2

f(1) = 2 + 1 – 5 + 2 = 0 

f(1) = 0,  

Hence, x = 1 is also a zero of the given polynomial.

When x = -2 

f(-2) = 2(-2)3 + (-2)2 – 5(-2) + 2 

f(-2) = -16 + 4 + 10 + 2 = 0 

f(-2) = 0,   

Hence, x = -2 is also a zero of the given polynomial.

We know that Sum of zeros = -b/a;

Sum of the products of the zeros taken two at a time = c/a; 

Product of zeros = -d/a;

Here sum=1/2 + 1 – 2 =-1/2 and -b/a=-1/2

Here sum of products=(1/2 * 1) + (1 * -2) + (1/2 * -2) =-5/2 and c/a=-5/2

Here product =1/2 x 1 x (- 2) = -1 and -d/a=-1



Hence, the relationship between the zeros and coefficients is verified.

(ii) g(x) = x3 – 4x2 + 5x – 2; 2, 1, 1

Solution: 

For the followings numbers to be the zeros of polynomial they must satisfy the following equation ie. g(x)=0 so now checking:

When x = 2 

g(2) = (2)3 – 4(2)2 + 5(2) – 2 

g(2) = 8 – 16 + 10 – 2 = 0 

g(2) = 0, 

Hence, x = 2 is a zero of the given polynomial.

Now we have two same roots, so we will check only once.

When x = 1 

g(1) = (1)3 – 4(1)2 + 5(1) – 2 

g(1) = 1 – 4 + 5 – 2 = 0 

g(1) = 0, 

Hence, x = 1 is also a zero of the given polynomial.

We know that Sum of zeros = -b/a;

Sum of the products of the zeros taken two at a time = c/a; 

Product of zeros = – d/a;

Here sum=2+1+1= 4 and -b/a=4 

Here sum of products=(1 * 1) + (1 * 2) + (2 * 1) =5 and c/a =5

Here product = 2*1*1=2 and -d/a=2

Hence, the relationship between the zeros and coefficients is verified.

Question 2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and product of its zeros as 3, -1, and -3 respectively.

Solution: 

A cubic polynomial say, f(x) is of the form ax3 + bx2 + cx + d.

We know that f(x) = k [x3 – (sum of roots)x2 + (sum of products of roots taken two at a time)x -(product of roots)]

Sum of roots = 3; 

Sum of products of roots taken two at a time=-1; 

Product of roots=-3

f(x) = k [x3 – (3)x2 + (-1)x – (-3)]

∴ f(x) = k [x3 – 3x2 – x + 3)]

Required polynomial f(x) = k [x3 – (3)x2 + (-1)x – (-3)]

∴ f(x) = k[x3 – 3x2 – x + 3)]

Question 3. If the zeros of the polynomial f(x) = 2x3 – 15x2 + 37x – 30 are in A.P., find them.

Solution:

 Let the roots be α = a – d, β = a and γ = a +d, Where, a is the first term and d is the common difference.

From given f(x), a= 2, b= -15, c= 37 and d= 30

=> Sum of roots = α + β + γ = (a – d) + a + (a + d) = 3a = (-b/a) = -(-15/2) = 15/2

 So, calculating for a, we get 3a = 15/2 i.e. a = 5/2

=> Product of roots = (a – d) x (a) x (a + d) = a(a2 –d2) = -d/a = -(30)/2 = 15 i.e. a(a2 –d2) = 15

Substituting ‘a’ we get ∴ d = 1/2 or -1/2

When d=1/2

Roots are α = 5/2-1/2, β = 5/2 and γ = 5/2 +1/2 i.e. α = 2, β = 2.5 and γ = 3

When d=-1/2

Roots are α = 5/2-(-1/2), β = 5/2 and γ = 5/2 +(-1/2) i.e. α = 3, β = 2.5 and γ = 2

Question 4. Find the condition that the zeros of polynomial f(x) = x3+3px2+3qx+r may be in A.P.

Solution:

 Let the roots be α = A – D, β = A and γ = A +D, Where, A is the first term and D is the common difference.

sum = A-D+A+A+D = -b/a i.e. 3A=-3b so, A=-b

Since β = A is a root so f(A)=0;

a3+3pa2+3qa+r=0

Now put a=-p; so, we get

2p2-3pq+r=0 is the required condition.

Question 5. If the zeros of the polynomial f(x) = ax3+3bx2+3cx+d are in A.P. Prove that 2b3-3abc+a2d = 0. 

Solution:

Let the roots be α = A – D, β = A and γ = A +D, Where, A is the first term and D is the common difference.

sum = A-D+A+A+D = -b/a i.e. 3A=-3b/a so, A=-b/a

Now f(A)=0 so,

f(x)=aA3+3bA2+3cA+d   

Now put A=-b/a; So, we get: 

2b3-3abc+a2d =0

Hence, proved.

Question 6. If the zeros of the polynomial f(x) = x3-12x2+39x+k are in A.P, find the value of k.

Solution:

Let the roots be α = A – D, β = A and γ = A +D, Where, A is the first term and D is the common difference.

sum = A-D+A+A+D = -b/a i.e., 3A=12 so, A=4

f(β)=0 i.e. f(A)=0

(4)3-12(4)2+39(4)+k=0

64-192+156+k=0

28+k=0

k=-28

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