RBSE Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 2 Polynomials Exercise 2.2

Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x² – 2x – 8
(ii) 4s² – 4s + 1
(iii) 6x² – 3 – 7x
(iv) 4u² + 8u
(v) t² – 15
(vi) 3x² – x – 4
Solution:
(i) According to the question
x² – 2x – 8 = x² + 2x – 4x – 8
= x (x + 2) – 4 (x + 2)
= (x + 2) (x – 4)
Therefore the value of x² – 2x – 8 will be zero if the value of (x + 2) (x – 4) is zero,
i. e., x + 2 = 0 or x – 4 = 0, i.e., x = – 2 or x = 4.
Therefore the zeroes of x² – 2x – 8 are – 2 and 4.
Now, Sum of the zeroes = (- 2) + 4 = 2

(ii) According to the question 4s² – 4s + 1 = 4s² – 2s – 2s + 1
= 2s (2s – 1) – 1 (2s – 1)
= (2s – 1) (2s – 1)
Therefore, the value of 4s² – 4s + 1 will be zero if the value of (2s – 1) (2s – 1) is zero,
i. e., 2s – 1 = 0 or 2s – 1 = 0,
i.e., s = \(\frac{1}{2}\) s = \(\frac{1}{2}\)
Hence the zeroes of 4s² – 4s + 1 are \(\frac{1}{2}\) and \(\frac{1}{2}\)
Now, sum of the zeroes

(iii) According to the question 6x² – 3 – 7x = 6x² – 7x – 3
= 6x² – 9x + 2x – 3
= 3x (2x – 3) + 1 (2x – 3)
= (3x + 1) (2x – 3)
So, the value of 6x² – 3 – 7x will be zero if the value of (3x + 1) (2x – 3) is zero, i.e.,
3x + 1 = 0 or 2x – 3 = 0
i.e., x = \(-\frac{1}{3}\) or x = \(\frac{3}{2}\)
Hence, the zeroes of 6x² – 3 – 7x are \(\frac{1}{3}\) and \(\frac{3}{2}\)
Now,sum of the zeroes = \(-\frac{1}{3}+\frac{3}{2}\)

(iv) According to the question 4u² + 8u = 4u (u + 2)
Therefore the value of 4u² + 8u will be zero if the value of 4u (u + 2) is zero, i.e.,
² = 0 or u + 2 = 0
i.e.. u = 0 or u = – 2
Hence, die zeroes of 4u² + 8M are 0 and – 2.
Now,
Sum of the zeroes = 0 + (- 2)

(v) According to the question
t² – 15 = (t – \(\sqrt{15}\)) (t + \(\sqrt{15}\))
Therefore the value of t² – 15 will be zero if the value of (t – \(\sqrt{15}\)) (t + \(\sqrt{15}\)) is zero.
i.e.. t – \(\sqrt{15}\) = 0 or t + \(\sqrt{15}\) = 0
i.e., t= \(\sqrt{15}\) or t = \(-\sqrt{15}\)
Hence, the zeroes of t² – 15 are \(\sqrt{15}\) and \(-\sqrt{15}\)
Now, sum of the zeroes = \(\sqrt{15}\) + (\(-\sqrt{15}\))
= 0 = \(\frac{0}{1}\)

(vi) According to the question 3x² – x – 4 = 3x² + 3x – 4x – 4
= 3²(x + 1) – 4(x + 1)
= (x + 1) (3x – 4)
The value of 3x² – x – 4 will be zero if the value of (x + 1) (3x – 4) is zero,
i.e., x + 1 = 0 or 3x – 4 = 0
i.e., x = – 1 or x = \(\frac{4}{3}\)
Hence, the zeroes of 3x² – x – 4 are – 1 and \(\frac{4}{3}\)
Now, sum of the zeroes = – 1 + \(\frac{4}{3}\)

Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

Solution:
(i) Let the polynomial be ax² + bx + c and its zeroes be α and β.
Then, α + β = \(\frac{1}{4}\) = \(\frac{-b}{a}\)
and αβ = – 1 = \(\frac{-4}{4}\) = \(\frac{c}{a}\)
If a = 4, b = – 1 and c = – 4
Hence the quadratic polynomial will be
4x² – x – 4.

(ii) Let the polynomial be ax² + bx + c and its zeroes be α and β.
Then α + β = \(\sqrt{2}\) = \(\frac{3 \sqrt{2}}{3}\) = \(\frac{-b}{a}\)
and αβ = \(\frac{1}{3}\) = \(\frac{c}{a}\)
If a = 3, b = – 3\(\sqrt{2}\) and c = 1
Hence, the quadratic polynomial will be
3x² – 3\(\sqrt{2}\)x + 1

(iii) Let the polynomial be ax² + bx + c and its zeroes be α and β.
Then α + β = 0 = \(\frac{0}{1}\) = \(\frac{-b}{a}\)
and, αβ = \(\sqrt{5}\) = \(\frac{\sqrt{5}}{1}\) = \(\frac{c}{a}\)
If a = 1, b = 0 and c = \(\sqrt{5}\)
Hence, the quadratic polynomial will be x² – 0.x + \(\sqrt{5}\) or x² + \(\sqrt{5}\)

(iv) Let the polynomial be ax² + bx + c and its zeroes be α and β.
Then, α + β = 1 = \(\frac{-(-1)}{1}\) = \(\frac{-b}{a}\)
and, αβ = 1 = \(\frac{1}{1}\) = \(\frac{c}{a}\)
If a = 1, b = – 1 and c = 1
Hence, the quadratic polynomial will be x² – x + 1.

(v) Let the polynomial be ax² + bx + c and its zeroes be α and β.
Then. α + β = \(-\frac{1}{4}\) = \(\frac{-b}{a}\)
and, αβ = \(\frac{1}{4}\) = \(\frac{c}{a}\)
If a = 4 b = 1 and c = 1.
Hence, the quadratic polynomial will be 4x² + x + 1

(vi) Let the polynomial be ax² + bx + c and its zeroes be α and β.
Then, α + β = 4 = \(\frac{-b}{a}\)
and, αβ = 1 = \(\frac{1}{1}\) = \(\frac{c}{a}\)
If a = 1, b = – 4 and c = 1.
Hence, the quadratic polynomial will be x² – 4x + 1.