Class 10 Mathematics

# Polynomials

## Exercise 2.2 (NCERT)

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

Solution: x^2 – 2x – 8 = 0
Or, x^2 – 4x + 2x – 8 = 0
Or, x(x – 4) + 2(x – 4) = 0
Or, (x + 2)(x – 4) = 0

Hence, zeroes are -2 and 4

We know that; sum of zeroes = - b/a
Or, - 2 + 4 = - (-2)
Or, text(LHS) = text(RHS)

Again we know that; product of zeroes = c/a

Or, - 2 xx 4 = - 8

From sum and product of zeroes, the relationship between the zeroes and coefficients is verified.

(ii) 4s^2 – 4s + 1

Solution: 4s^2 – 4s + 1 = 0
Or, 4s^2 – 2s – 2s + 1 = 0
Or, 4s(s – ½ ) – 2(s – ½ ) = 0
Or, (4s – 2)(s – ½) = 0

Here, zeroes are; ½

We know that; sum of zeroes = - b/a
Or, ½ + ½ = - (- 4/4)
Or, 1 = 1

We know that; product of zeroes = c/a
Or, ½ xx ½ = ¼

From sum and product of zeroes, the relationship between the zeroes and coefficients is verified.

(iii) 6x^2 – 3 – 7x

Solution: 6x^2 – 7x – 3 = 0
Or, 6x^2 - 9x + 2x – 3 = 0
Or, 3x (2x – 3) + 1(2x – 3) = 0
Or, (3x + 1) (2x – 3) = 0

Here, zeroes are; - ½ and 3/2

We know that; sum of zeroes = - b/a
Or, 1/2+1/2=-(-4/4)
Or, 1=1

We know that products of zeroes =c/a
Or, 1/2xx1/2=1/4

From sum and product of zeroes, the relationship between the zeroes and coefficients is verified.

(iv) 4u^2 + 8u

Solution: 4u^2 + 8u = 0

Or, u^2 + 2u = 0

Or, u(u + 2) = 0

Hence, zeroes are; 0 and – 2

We know that; sum of zeroes = - b/a

Or, 0 – 2 = - 2

We know that; product of zeroes = c/a

Or, 0 xx (- 2) = 0

From sum and product of zeroes, the relationship between the zeroes and coefficients is verified

(v) t^2 – 15

Solution: t^2 – 15 = 0
Or, t^2 = 15
Or, t = sqrt15

Hence, zeroes are ±sqrt15

We know that; sum of zeroes = - b/a

Or, - sqrt15 + sqrt15 = 0

We know that; product of zeroes = c/a

Or, sqrt15 xx\ sqrt15 = 15

From sum and product of zeroes, the relationship between the zeroes and coefficients is verified.

(vi) 3x^2 – x – 4

Solution: 3x^2 – x – 4 = 0
Or, 3x^2 + 3x – 4x – 4 = 0
Or, 3x(x + 1) – 4(x + 1) = 0
Or, (3x – 4)(x + 1) = 0

Hence, zeroes are; 4/3 and – 1

We know that; sum of zeroes = - b/a

Or, 4/3-1=1/3

Or, (4-3)/(3)=1/3

We know that; product of zeroes = c/a

Or, 4/3xx(-1)=-4/3

From sum and product of zeroes, the relationship between the zeroes and coefficients is verified.

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

(i) ¼ , -1

Solution: We know that a quadratic equation can be given as follows:

x^2 – text((sum of zeroes))x + text(product of zeroes)

Hence; the required equation can be written as follows:

x^2-1/4x-1

=4x^2-x-4

(ii) sqrt2, 1/3

Solution: We know that a quadratic equation can be given as follows:

x^2 – text((sum of zeroes))x + text(product of zeroes)

Hence; the required equation can be written as follows:

x^2-sqrt2x+1/3

=3x^2-3sqrt2x+1

(iii) 0, sqrt5

Solution: We know that a quadratic equation can be given as follows:

x^2 – text((sum of zeroes))x + text(product of zeroes)

Hence; the required equation can be written as follows:

x^2-9x+sqrt5

=x^2+sqrt5

(iv) 1, 1

Solution: We know that a quadratic equation can be given as follows:

x^2 – text((sum of zeroes))x + text(product of zeroes)

Hence; the required equation can be written as follows:

x^2 – x + 1

(v) – ¼, ¼

Solution: We know that a quadratic equation can be given as follows:

x^2 – text((sum of zeroes))x + text(product of zeroes)

Hence; the required equation can be written as follows:

x^2+1/4x+1/4

=4x^2+x+1

(vi) 4, 1

Solution: We know that a quadratic equation can be given as follows:

x^2 – text((sum of zeroes))x + text(product of zeroes)

Hence; the required equation can be written as follows:

x^2 – 4x + 1

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Exercise 2.1

Exercise 2.3