KSEEB SSLC Class 10 Maths Solutions Chapter 9 Polynomials Ex 9.2

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) x² – 2x – 8
(ii) 4s² – 4s + 1
(iii) 6x² – 3 – 7x
(iv) 4u² – 8u
(v) t² – 15
(vi) 3x² – x – 4
Solution:

(iv) 4u² – 8u
= 4u² – 8u + 0
= 4u (u – 2)
If 4u = 0, then u = 0
If u – 2 = 0, then u = 2
∴ Zeroes are 0 and 2

(v) t² – 15
= t² + 0 – 15

(vi) 3x² – x – 4
= 3x² – 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(x² – sum of the zeroes) x + product of zeroes.
= K(x² – 1x + 1)
Taking K = 1
= x² – 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[x² – (sum of the zeroes) x + Product of zeroes]
= K(x² – 4x + 1)
Taking K = 1
= 1(x² – 4x + 1)
= x² – 4x + 1

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