RD Sharma Class 10 Solutions Chapter 2 Exercise 2.2: This exercise focuses on verifying the zeros and the relationship between the zeros and the coefficients of a cubic polynomial. The experts at Kopykitab have prepared RD Sharma Solutions Class 10 to help students understand these concepts clearly while following the latest CBSE patterns. For a detailed explanation of the questions in this exercise, students can get RD Sharma Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2 PDF from the link below.
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RD Sharma Class 10 Solutions Chapter 2 Exercise 2.2
Access answers to RD Sharma Solutions Class 10 Maths Chapter 2 Exercise 2.2- Important Question with Answers
1. 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:
Given, f(x) = 2x3 + x2 – 5x + 2, where a= 2, b= 1, c= -5 and d= 2
For x = 1/2
f(1/2) = 2(1/2)3 + (1/2)2 – 5(1/2) + 2
= 1/4 + 1/4 – 5/2 + 2 = 0
⇒ f(1/2) = 0, hence x = 1/2 is a root of the given polynomial.
For x = 1
f(1) = 2(1)3 + (1)2 – 5(1) + 2
= 2 + 1 – 5 + 2 = 0
⇒ f(1) = 0, hence x = 1 is also a root of the given polynomial.
For x = -2
f(-2) = 2(-2)3 + (-2)2 – 5(-2) + 2
= -16 + 4 + 10 + 2 = 0
⇒ f(-2) = 0, hence x = -2 is also a root of the given polynomial.
Now,
Sum of zeros = -b/a
1/2 + 1 – 2 = – (1)/2
-1/2 = -1/2
Some of the products of the zeros taken two at a time = c/a
(1/2 x 1) + (1 x -2) + (1/2 x -2) = -5/ 2
1/2 – 2 + (-1) = -5/2
-5/2 = -5/2
Product of zeros = – d/a
1/2 x 1 x (– 2) = -(2)/2
-1 = -1
Hence, the relationship between the zeros and coefficients is verified.
(ii) g(x) = x3 – 4x2 + 5x – 2; 2, 1, 1
Solution:
Given, g(x) = x3 – 4x2 + 5x – 2, where a= 1, b= -4, c= 5 and d= -2
For x = 2
g(2) = (2)3 – 4(2)2 + 5(2) – 2
= 8 – 16 + 10 – 2 = 0
⇒ f(2) = 0, hence x = 2 is a root of the given polynomial.
For x = 1
g(1) = (1)3 – 4(1)2 + 5(1) – 2
= 1 – 4 + 5 – 2 = 0
⇒ g(1) = 0, hence x = 1 is also a root of the given polynomial.
Now,
Sum of zeros = -b/a
1 + 1 + 2 = – (-4)/1
4 = 4
Some of the products of the zeros taken two at a time = c/a
(1 x 1) + (1 x 2) + (2 x 1) = 5/ 1
1 + 2 + 2 = 5
5 = 5
Product of zeros = – d/a
1 x 1 x 2 = -(-2)/1
2 = 2
Hence, the relationship between the zeros and coefficients is verified.
2. Find a cubic polynomial with the sum, the sum of the product of its zeroes taken two at a time, and the product of its zeros as 3, -1, and -3 respectively.
Solution:
Generally,
A cubic polynomial say f(x) is of the form ax3 + bx2 + cx + d.
And, can be shown w.r.t its relationship between roots as.
⇒ f(x) = k [x3 – (sum of roots)x2 + (sum of products of roots taken two at a time)x – (product of roots)]
Where k is any non-zero real number.
Here,
f(x) = k [x3 – (3)x2 + (-1)x – (-3)]
∴ f(x) = k [x3 – 3x2 – x + 3)]
where k is any non-zero real number.
3. If the zeros of the polynomial f(x) = 2x3 – 15x2 + 37x – 30 are in A.P., find them.
Solution:
Let the zeros of the given polynomial be α, β and γ. (3 zeros as it’s a cubic polynomial)
And given, the zeros are in A.P.
So, let’s consider the roots as
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 ⇒ a = 5/2
⇒ Product of roots = (a – d) x (a) x (a + d) = a(a2 –d2) = -d/a = -(30)/2 = 15
⇒ a(a2 –d2) = 15
Substituting the value of a, we get
⇒ (5/2)[(5/2)2 –d2] = 15
⇒ 5[(25/4) –d2] = 30
⇒ (25/4) – d2 = 6
⇒ 25 – 4d2 = 24
⇒ 1 = 4d2
∴ d = 1/2 or -1/2
Taking d = 1/2 and a = 5/2
We get,
the zeros as 2, 5/2 and 3
Taking d = -1/2 and a = 5/2
We get,
the zeros as 3, 5/2 and 2
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