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Question Number 1041 by tera last updated on 22/May/15

i f t h e r o o t s o f t h e e q u a t i o n {a x}^{2} + 5 x - 2 = 0 i s 2 a n d b .

t h e n 4 a^{2} - 4 a b + b^{2} = \dots

Answered by prakash jain last updated on 23/May/15

Since x = 2 is a root (x - 2) is a factor of

{a x}^{2} + 5 x - 2

Put x = 2

4 a + 10 - 2 = 0 \Rightarrow a = - 2

- 2 x^{2} + 5 x - 2 = 0

- 2 x^{2} + 4 x + x - 2 = 0

- (2 x - 1) (x - 2) = 0

Roots at 2 and \frac{1}{2}, hence b = \frac{1}{2}

4 a^{2} - 4 a b + b^{2} = 4 {(- 2)}^{2} - 4 (- 2) (\frac{1}{2}) + {(\frac{1}{2})}^{2}

= 16 + 4 + \frac{1}{4} = \frac{81}{4}

Commented by tera last updated on 23/May/15

t h a n k y o u s i r

Answered by prakash jain last updated on 22/May/15

s u m o f r o o t s 2 + b = \frac{- 5}{a} \Rightarrow 2 a + a b = - 5

p r o d u c t o f r o o t s 2 b = \frac{- 2}{a} \Rightarrow a b = - 1

2 a + a b = - 5 \Rightarrow 2 a - 1 = - 5 \Rightarrow a = - 2

a b = - 1 \Rightarrow b = \frac{1}{2}

p u t v a l u e s i n 4 a^{2} - 4 a b + b^{2}