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## Karnataka State Syllabus Class 10 Maths Chapter 9 Polynomials Ex 9.2

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

Answer:

(i) x^{2} – 2x – 8

P(x) = x^{2} – 2x – 8

P(x) = x^{2} – 4x + 2x – 8

P(x) = x(x – 4) + 2(x – 4)

(x – 4)(x + 2) = 0

x = 4(or) x = – 2

x = 4, – 2

P(x) = 0 zero of the polynomial P(x)

∴ Zeroes of the polynomial P (x) are 4 and – 2.

Sum of zeroes = 4 + (- 2) = – \(\frac{(2)}{1}\)

Product of zeroes = 4 × – 2 = – 8

(ii) 4s^{2} – 4s + 1

P(s) = 4s^{2} – 4s + 1

= 4s^{2} – 2s – 2s + 1

P(s) = 2s(2s – 1) – 1(2s – 1)

P(s) = 0 zero of the polynomial P(s)

(2s – 1)(2s – 1) = 0

2s – 1 = 0 (or) 2s – 1 = 0

2s = 1 (or) 2s = 1

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

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

∴ Zeroes of polynomial P(s) are 1/2 and 1/2

(iii) 6x^{2} – 3 – 7x

P(x) = 6x^{2} – 7x – 3

= 6x^{2} – 9x + 2x – 3

P(x) = 3x(2x – 3) + 1(2x – 3)

P(x) = (2x – 3)(3x + l)

P(x) = 0 zero of the polynomial P(x)

(2x – 3)(3x + 1) = 0

2x – 3 = 0 (or) 3x + 1 =0

2x = 3 (or) 3x = – 1

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

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

P(u) = 4u(u + 2)

P(u) = 0 zero of the polynomial P(u)

4u(u + 2) = 0

4u = 0 (or) u + 2 = 0

u = \(\frac{0}{4}\) (or) u = – 2

u = 0 (or) u = – 2

u = 0, – 2

∴ zeroes of polynomial P(u) are 0, – 2

sum of zeroes = 0 + ( – 2) = – 2

Product of zeroes = 0 × – 2 = 0

(v) t^{2} – 15

P(t) = t^{2} – 15

P(t) = t^{2} – (\(\sqrt{15}\))^{2}

P(t) = 0, zero of the polynomial P(t)

(vi) 3x^{2} – x – 4

3x^{2} – x – 4

P(x) = 3x^{2} – x – 4

P(x) = 3x^{2} – 4x + 3x – 4

= x(3x – 4) + 1(3x – 4)

P(x) = (3x – 4)(x + 1)

P(x) = 0, zero of the polynomial P(x)

(3x – 4)(x + 1) = 0

3x – 4 = 0 (or) x + 1 = 0

3x = 4 (or) x = – 1

x = \(\frac{4}{3}\) (or) x = – 1

x = \(\frac{4}{3}\), – 1

zeroes of the polynomial P(x) are \(\frac{4}{3}\), – 1

sum of zeroes = \(\frac{4}{3}\) + (- 1)

Question 2.

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

(i) \(\frac{1}{4}\), – 1

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

(iii) 0, \(\sqrt{5}\)

(iv) 1, 1

(v) – \(\frac{1}{4}\), \(\frac{1}{4}\)

(vi) 4, 1

Answer:

(i) Standard form of Quadratic polynomial

sum and product of its zeroes is

K[x^{2} – (sum of the zeroes) x + product of zeroes]

= K(x^{2} – \(\frac{1}{4}\)x – 1)

= 4x^{2} – x – 4

(ii) Standard form of quadratic polynomial sum and product of its zeroes is.

K [x^{2} – (sum of the zeroes) x + product of zeroes]

(iii) Standard form of quadratic polynomial sum and product of its zeroes is.

K[x^{2} – (sum of the zeroes ) x + product of zeroes]

= K(x^{2} – 0x + \(\sqrt{5}\)) = K(x^{2} + \(\sqrt{5}\))

Taking K = 1

= x^{2} + \(\sqrt{5}\)

(iv) 1, 1

Here m + n = 1, mn = 1

∴ Quadratic Equation

x^{2} – (m + n)x + mn

x^{2} – (1)x + 1

x^{2} – x + 1

(v) Standard form of quadratic polynomial sum and product of its zeroes is.

K [x^{2} – (sunt of the zeroes) x + product of zeroes]

= 4x^{2} – x + 1

(vi) 4, 1

Here, m + n = 4, mn = 1

∴ Quadratic Equation

x^{2} – (m + n)x + mn

x^{2} – (4)x + 1

x^{2} – 4x + 1