KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.2

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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) x² – 2x – 8
(ii) 4s² – 4s + 1
(iii) 6x² – 3 – 7x
(iv) 4u² + 8u
(v) t² – 15
(vi) 3x² – x – 4
Answer:
(i) x² – 2x – 8
P(x) = x² – 2x – 8
P(x) = x² – 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) 4s² – 4s + 1
P(s) = 4s² – 4s + 1
= 4s² – 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) 6x² – 3 – 7x
P(x) = 6x² – 7x – 3
= 6x² – 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) 4u² + 8u
P(u) = 4u² + 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) t² – 15
P(t) = t² – 15
P(t) = t² – (\(\sqrt{15}\))²
P(t) = 0, zero of the polynomial P(t)

(vi) 3x² – x – 4
3x² – x – 4
P(x) = 3x² – x – 4
P(x) = 3x² – 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[x² – (sum of the zeroes) x + product of zeroes]
= K(x² – \(\frac{1}{4}\)x – 1)

= 4x² – x – 4

(ii) Standard form of quadratic polynomial sum and product of its zeroes is.
K [x² – (sum of the zeroes) x + product of zeroes]

(iii) Standard form of quadratic polynomial sum and product of its zeroes is.
K[x² – (sum of the zeroes ) x + product of zeroes]
= K(x² – 0x + \(\sqrt{5}\)) = K(x² + \(\sqrt{5}\))
Taking K = 1
= x² + \(\sqrt{5}\)

(iv) 1, 1
Here m + n = 1, mn = 1
∴ Quadratic Equation
x² – (m + n)x + mn
x² – (1)x + 1
x² – x + 1

(v) Standard form of quadratic polynomial sum and product of its zeroes is.
K [x² – (sunt of the zeroes) x + product of zeroes]

= 4x² – x + 1

(vi) 4, 1
Here, m + n = 4, mn = 1
∴ Quadratic Equation
x² – (m + n)x + mn
x² – (4)x + 1
x² – 4x + 1