Jharkhand Board JAC Class 10 Maths Solutions Chapter 2 Polynomials Ex 2.2 Textbook Exercise Questions and Answers.

## JAC Board 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 :

1. x^{2} – 2x – 8

2. 4s^{2} – 4s + 1

3. 6x^{2} – 3 – 7x

4. 4u^{2} + 8u

5. t^{2} – 15

6. 3x^{2} – x – 4

Solution:

1. x^{2} – 2x – 8 = x^{2} – 4x + 2x – 8

= x(x – 4) + 2(x – 4)

= (x – 4)(x + 2)

So, the value of x^{2} – 2x – 8 is zero, when

x – 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = -2.

Hence, the zeroes of polynomial x^{2} – 2x – 8 are 4 and -2.

Now,

Sum of zeroes = 4 + (-2)

= 2

= \(\frac{-(-2)}{1}\)

= \(\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^2}\)

and Product of zeroes = (4) (-2)

= -8

= \(\frac{-8}{1}\)

= \(\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)

2. 4s^{2} – 4s + 1 = 4s^{2} – 2s – 2s + 1

= 2s (2s – 1) -1 (2s – 1)

= (2s – 1)(2s – 1)

So, the value of 4s^{2} – 4s + 1 is zero, when \(\frac{1}{2}\) or

s = \(\frac{1}{2}\)

2s – 1 = 0 or 2s – 1 = 0, i.e., when s = 2

Hence, the zeroes of polynomial 4s^{2} – 4s + 1 are \(\frac{1}{2}\) and \(\frac{1}{2}\) (both equal).

Now,

3. 6x^{2} – 3 – 7x = 6x^{2} – 9x + 2x – 3

= 3x(2x – 3) + 1(2x – 3)

= (2x – 3)(3x + 1)

So, 6x^{2} – 3 – 7x = 0 when 2x – 3 = 0 or 3x + 1 = 0,

i.e., when x = \(\frac{3}{2}\) or x = –\(\frac{1}{3}\).

Hence, the zeroes of polynomial 6x^{2} – 3 – 7x are \(\frac{3}{2}\) and –\(\frac{1}{3}\).

Now,

4. 4u^{2} + 8u = 4u (u + 2)

So, 4u^{2} + 8u = 0 when 4u = 0 or u + 2 = 0,

i.e., when u = 0 or u = -2.

Hence, the zeroes of polynomial 4u^{2} + 8u are 0 and -2.

Now,

Sum of zeroes = 0 + (-2)

= -2

= \(\frac{-(8)}{4}\)

= \(\frac{-(\text { Coefficient of } u)}{\text { Coefficient of } u^2}\)

and

Product of zeroes = (0)(-2) = 0

= \(\frac{0}{4}\)

= \(\frac{\text { Constant term }}{\text { Coefficient of } u^2}\)

Note: In polynomial

4u^{2} + 8u = 4u^{2} + 8u + 0, the constant term is 0.

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

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

So, t^{2} – 15 = 0 when t + \(\sqrt{15}\) = 0 or

t – 15 = 0, i.e., when t = –\(\sqrt{15}\) or t = \(\sqrt{15}\).

Hence, the zeroes of polynomial t^{2} – 15 are –\(\sqrt{15}\) and \(\sqrt{15}\).

Now,

Sum of zeroes = (\(-\sqrt{15}\)) + (\(\sqrt{15}\))

= 0

= \(\frac{-0}{1}\)

= \(\frac{-(\text { Coefficient of } t)}{\text { Coefficient of } t^2}\)

and

Product of zeroes = (\(\sqrt{15}\)) (\(\sqrt{15}\))

= -15

= \(\frac{-15}{1}\)

= \(\frac{\text { Constant term }}{\text { Coefficient of } t^2}\)

Note: In polynomial t^{2} – 15 = t^{2} + 0t – 15, the coefficient of t is 0.

6. 3x^{2} – x – 4 = 3x^{2} + 3x – 4x – 4

= 3x(x + 1) – 4(x + 1)

= (x + 1)(3x – 4)

So, 3x^{2} – x – 4 = 0 when x + 1 = 0 or 3x – 4 = 0, i.e., when x = -1 or x = \(\frac{4}{3}\)

Hence. -1 and \(\frac{4}{3}\) are the zeroes of polynomial 3x^{2} – x – 4.

Now,

Question 2.

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

1. \(\frac{1}{4}\), -1

2. \(\sqrt{2}\), \(\frac{1}{3}\)

3. 0, \(\sqrt{5}\)

4. 1, 1

5. \(-\frac{1}{4}, \frac{1}{4}\)

6. 4, 1

Solution:

1. \(\frac{1}{4}\), -1

Let the quadratic polynomial be ax^{2} + bx + c and its zeroes be α and β.

Then, as per given,

α + β = \(\frac{1}{4}\) = \(\frac{-b}{a}\)

and αβ = -1 = \(\frac{c}{a}\)

If a = 4, then b = -1 and c = -4.

So, one quadratic polynomial satisfying the given condition is 4x^{2} – x – 4.

In general, a quadratic polynomial satisfying the given condition is k(4x^{2} – x – 4), where k is a non-zero real number.

2. \(\sqrt{2}\), \(\frac{1}{3}\)

Let the quadratic polynomial be ax^{2} + bx + c and its zeroes be α and β.

Then, as per given,

α + β = \(\sqrt{2}\) = \(\frac{-b}{a}\)

and αβ = \(\frac{1}{3}\) = \(\frac{c}{a}\)

If a = 3, then b = -3\(\sqrt{2}\) and c = 1

So, one quadratic polynomial satisfying the given condition is 3x^{2} – 3\(\sqrt{2}\)x + 1.

In general, a quadratic polynomial satisfying the given condition is k(3x^{2} – 3\(\sqrt{2}\)x + 1).

where k is a non-zero real number.

3. 0, \(\sqrt{5}\)

Let the quadratic polynomial be ax^{2} + bx + c and its zeroes be α and β.

Then, as per given, α + β = 0 = \(\frac{-b}{a}\)

and αβ = \(\sqrt{5}\) = \(\frac{c}{a}\)

If a = 1, then b = 0 and c = \(\sqrt{5}\).

So, one quadratic polynomial satisfying the given condition is x^{2} + \(\sqrt{5}\).

In general, a quadratic polynomial satisfying the given condition is k(x^{2} + \(\sqrt{5}\)), where k is a non-zero real number.

4. 1, 1

Let the quadratic polynomial be ax^{2} + bx + c and its zeroes be α and β.

Then, as per given.

α + β = 1 = \(\frac{-b}{a}\)

and αβ = 1 = \(\frac{c}{a}\)

If a = 1, then b = -1 and c = 1.

So, one quadratic polynomial satisfying the given condition is x^{2} – x + 1.

In general, a quadratic polynomial satisfying the given condition is k(x^{2} – x + 1), where k is a non-zero real number.

5. \(-\frac{1}{4}, \frac{1}{4}\)

Let the quadratic polynomial be ax^{2} + bx + c and its zeroes be α and β.

Then, as per given,

α + β = \(-\frac{1}{4}\) = \(\frac{-b}{a}\)

and αβ = \(\frac{1}{4}\) = \(\frac{c}{a}\)

If a = 4, then b = 1 and c = 1.

So, one quadratic polynomial satisfying the given condition is 4x^{2} + x + 1.

In general, a quadratic polynomial satisfying the given condition is k(4x^{2} + x + 1), where k is a non-zero real number.

6. 4, 1

Let the quadratic polynomial be ax^{2} + bx + c and its zeroes be α and β.

Then, as per given.

α + β = 4 = \(\frac{-b}{a}\)

and αβ = 1 = \(\frac{c}{a}\)

If a = 1, then b = -4 and c = 1.

So, one quadratic polynomial satisfying the given condition is x^{2} – 4x + 1.

In general, a quadratic polynomial satisfying the given condition is k(x^{2} – 4x + 1), where k is a non-zero real number.