**KSEEB SSLC Class 10 Maths Solutions Chapter 9 Polynomials Ex 9.2** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 9.1.

## Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 9.2

Question 1.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and their 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:

(iv) 4u^{2} – 8u

= 4u^{2} – 8u + 0

= 4u (u – 2)

If 4u = 0, then u = 0

If u – 2 = 0, then u = 2

∴ Zeroes are 0 and 2

(v) t^{2} – 15

= t^{2} + 0 – 15

(vi) 3x^{2} – x – 4

= 3x^{2} – 4x + 3x – 4

= x(3x – 4) + 1 (3x – 4)

= (3x – 4) (x + 1)

If 3x – 4 = 0, then x = \(\frac{4}{3}\)

If x + 1 = 0, then x = -1

∴ Zeroes are \(\frac{4}{3}\) and -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

Solution:

(i) \(\frac{1}{4}\), 1

Here m + n = \(\frac{1}{4}\), mn = -1

∴ Quadratic Equation

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

Here m + n = \(\sqrt{2}\), mn = \(\frac{1}{3}\)

∴ Quadratic Equation

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

Here m + n = 0, mn = \(\sqrt{5}\)

∴ Quadratic Equation

(iv) 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} – 1x + 1)

Taking K = 1

= x^{2} – x + 1

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

Here m + n = \(-\frac{1}{4}\), mn = \(\frac{1}{4}\)

∴ Quadratic Equation

(vi) 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} – 4x + 1)

Taking K = 1

= 1(x^{2} – 4x + 1)

= x^{2} – 4x + 1

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