The equation x^2 +2xp+q=0 and x^2 +2ax+b=0 have common roots, show that (q−b)^2 +4(a−p)(aq−pb)=0


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Question Number 159653 by MathsFan last updated on 19/Nov/21

$T h e e q u a t i o n x^{2} + 2 x p + q = 0$ $a n d x^{2} + 2 a x + b = 0 h a v e c o m m o n$ $r o o t s, s h o w t h a t {(q - b)}^{2} + 4 (a - p) (a q - p b) = 0$
	Answered by Rasheed.Sindhi last updated on 19/Nov/21

	$T h e e q u a t i o n s h a v e c o m m o n r o o t s$ $\Rightarrow T {h e y}^{'} r e e q u i v a l e n t :$ $\frac{2 p}{2 a} = \frac{q}{b}$ $\Rightarrow \frac{p}{a} = \frac{q}{b} \Rightarrow p = \frac{a q}{b}$ ${(q - b)}^{2} + 4 (a - p) (a q - p b)$ $= {(q - b)}^{2} + 4 (a - \frac{a q}{b}) (a q - b (\frac{a q}{b}))$ $= {(q - b)}^{2} + 4 (\frac{a b - a q}{b}) (\frac{a b q - a b q}{b})$ $= {(q - b)}^{2} \neq 0 i n g e n e r a l .$ $O n l y e q u a l t o 0 w h e n q = b$
	Commented by MathsFan last updated on 19/Nov/21

	$t h a n k y o u s i r$
	Commented by mr W last updated on 19/Nov/21

	$p l e a s e r e c h e c k s i r :$ $i t m u s t b e q = b! o t h e r w i s e t h e r e a r e$ $n o c o m m o n r o o t s!$ $\Rightarrow T {h e y}^{'} r e e q u i v a l e n t :$ $\frac{2 p}{2 a} = \frac{q}{b} = \frac{1 (c o e f . o f x^{2} t e r m)}{1}$ $\Rightarrow p = a a n d q = b$ $e x a m p l e :$ $x^{2} + 3 x + 4 = 0 a n d x^{2} + 6 x + 8 = 0 h a v e$ $n o c o m m o n r o o t s, e v e n \frac{3}{6} = \frac{4}{8} .$
	Commented by Rasheed.Sindhi last updated on 20/Nov/21

	$R i g h t S i r, T h a n X!$
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