Given that ax+b is a factor of x^2 +2x^2 −1 and −a is a root of x^2 +2x −1=0 show that the value of b is −1 or 3+2(√(2 .))


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Question Number 40510 by KMA last updated on 23/Jul/18

$G i v e n t h a t a x + b i s a f a c t o r o f x^{2}$ $+ 2 x^{2} - 1 a n d - a i s a r o o t o f x^{2} + 2 x$ $- 1 = 0 s h o w t h a t t h e v a l u e o f b i s$ $- 1 o r 3 + 2 \sqrt{2 .}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jul/18

	$p u t x = \frac{- b}{a} i n x^{2} + 2 x - 1 t h e n R = 0$ ${(\frac{- b}{a})}^{2} + 2 (\frac{- b}{a}) - 1 = 0$ $b^{2} - 2 a b - a^{2} = 0$ $b = \frac{- (- 2 a) \pm \sqrt{4 a^{2} + 4 a^{2}}}{2}$ $b = a \pm . a \sqrt{2}$ $p u t x = (- a) i n x^{2} + 2 x - 1 = 0$ $a^{2} - 2 a - 1 = 0$ $a = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$ $b = a + a \sqrt{2} a n d a - a \sqrt{2}$ $b = 1 + \sqrt{2} + \sqrt{2} (1 + \sqrt{2}) = 3 + 2 \sqrt{2}$ $b = 1 - \sqrt{2} + \sqrt{2} (1 - \sqrt{2}) = 1 - \sqrt{2} + \sqrt{2} - 2 = - 1$
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