# KSEEB SSLC Class 10 Maths Solutions Chapter 9 Polynomials Ex 9.2

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### Karnataka State Syllabus Class 10 Maths SolutionsChapter 9 Polynomials Ex 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 = 43
If x + 1 = 0, then x = -1
∴ Zeroes are 43 and -1. Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 14, 1
(ii) 2–√,13
(iii) 0, 5–√
(iv) 1, 1
(v) −14,14
(vi) 4, 1
Solution:
(i) 14, 1
Here m + n = 14, mn = -1 (ii) 2–√,13
Here m + n = 2–√, mn = 13 (iii) 0, 5–√
Here m + n = 0, mn = 5–√ (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) −14,14
Here m + n = −14, mn = 14 (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

All Chapter KSEEB Solutions For Class 10 Maths

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All Subject KSEEB Solutions For Class 10

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