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) x^{2} – 2x – 8

(ii) 4s^{2} – 4s + 1

(iii) 6x^{2} – 3 – 7x

(iv) 4u^{2} + 8u

(v) t^{2} – 15

(vi) 3x^{2} – x – 4

Solution:

(i) According to the question

x^{2} – 2x – 8 = x^{2} + 2x – 4x – 8

= x (x + 2) – 4 (x + 2)

= (x + 2) (x – 4)

Therefore the value of x^{2} – 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 x^{2} – 2x – 8 are – 2 and 4.

Now, Sum of the zeroes = (- 2) + 4 = 2

(ii) According to the question 4s^{2} – 4s + 1 = 4s^{2} – 2s – 2s + 1

= 2s (2s – 1) – 1 (2s – 1)

= (2s – 1) (2s – 1)

Therefore, the value of 4s^{2} – 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 4s^{2} – 4s + 1 are \(\frac{1}{2}\) and \(\frac{1}{2}\)

Now, sum of the zeroes

(iii) According to the question 6x^{2} – 3 – 7x = 6x^{2} – 7x – 3

= 6x^{2} – 9x + 2x – 3

= 3x (2x – 3) + 1 (2x – 3)

= (3x + 1) (2x – 3)

So, the value of 6x^{2} – 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 6x^{2} – 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 4u^{2} + 8u = 4u (u + 2)

Therefore the value of 4u^{2} + 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 4u^{2} + 8M are 0 and – 2.

Now,

Sum of the zeroes = 0 + (- 2)

(v) According to the question

t^{2} – 15 = (t – \(\sqrt{15}\)) (t + \(\sqrt{15}\))

Therefore the value of t^{2} – 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 t^{2} – 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 3x^{2} – x – 4 = 3x^{2} + 3x – 4x – 4

= 3^{2}(x + 1) – 4(x + 1)

= (x + 1) (3x – 4)

The value of 3x^{2} – 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 3x^{2} – 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 ax^{2} + 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

4x^{2} – x – 4.

(ii) Let the polynomial be ax^{2} + 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

3x^{2} – 3\(\sqrt{2}\)x + 1

(iii) Let the polynomial be ax^{2} + 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 x^{2} – 0.x + \(\sqrt{5}\) or x^{2} + \(\sqrt{5}\)

(iv) Let the polynomial be ax^{2} + 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 x^{2} – x + 1.

(v) Let the polynomial be ax^{2} + 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 4x^{2} + x + 1

(vi) Let the polynomial be ax^{2} + 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 x^{2} – 4x + 1.

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