if the roots of 2x^2 −xn = 2x + m is 5, then find : 4n + m − 5


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Question Number 188327 by HeferH last updated on 28/Feb/23

$i f t h e r o o t s o f 2 x^{2} - x n = 2 x + m i s 5,$ $t h e n f i n d : 4 n + m - 5$
	Answered by Rasheed.Sindhi last updated on 28/Feb/23

	$2 x^{2} - x n = 2 x + m$ $2 x^{2} - x n - 2 x - m = 0$ $2 x^{2} - (n + 2) x - m = 0$ $S a y α, β a r e r o o t s$ $G i v e n α = β = 5$ $α + β = \frac{n + 2}{2} = 5 + 5 = 10 \Rightarrow n = 18$ $α β = - \frac{m}{2} = 5 \times 5 = 25 \Rightarrow m = - 50$ $4 n + m - 5 = 4 (18) + (- 50) - 5 = 17$
	Commented by HeferH last updated on 28/Feb/23

	$N i c e s o l u t i o n! I f o r g o r a b o u t t h o s e p r o p e r t i e s$ $o f r o o t s$
	Answered by Rasheed.Sindhi last updated on 28/Feb/23

	$A n o t h e r w a y . . .$ $2 x^{2} - x n = 2 x + m$ $2 x^{2} - x n - 2 x - m = 0$ $2 x^{2} - (n + 2) x - m = 0 . . . . (i)$ $T h e e q u a t i o n h a v i n g r o o t s 5, 5 i s$ ${(x - 5)}^{2} = 0$ $x^{2} - 10 x + 25 = 0 . . . . . . . . . (i i)$ $C o m p a r i n g c o e f f i c i e n t s o f (i) & (i i)$ $\frac{2}{1} = \frac{- (n + 2)}{- 10} = \frac{- m}{25}$ $n + 2 = 20 \land - m = 50$ $n = 18 \land m = - 50$ $▸ 4 n + m - 5 = 4 (18) + (- 50) - 5 = 17$
	Answered by Rasheed.Sindhi last updated on 28/Feb/23

	$2 x^{2} - (n + 2) x - m = 0$ $∙ ∵ 5 i s r o o t$ $∴ 2 {(5)}^{2} - (n + 2) (5) - m = 0$ $50 - 5 n - 10 - m = 0$ $m + 5 n = 40 . . . . . . . . . . (i)$ $∙ \frac{d}{d x} {(2 x^{2} - (n + 2) x - m)}_{n = 5} = \frac{d}{d x} {(0)}_{x = 5}$ ${4 x - (n + 2)]}_{x = 5} = 0$ $\Rightarrow 4 (5) - n - 2 = 0$ $\Rightarrow n = 18$ $(i) \Rightarrow m + 5 (18) = 40 \Rightarrow m = - 50$ $▸ 4 n + m - 5 = 4 (18) + (- 50) - 5 = 17$
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