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Question Number 7331 by rohit meena last updated on 23/Aug/16

Commented by sandy_suhendra last updated on 24/Aug/16

$f o r t h e s i m e t r i c r o o t, l i k e y_{1} = 2 α a n d y_{2} = 2 β, w e c a n u s e t h e s u b t i t u t e m e t h o d$ $y = 2 x \Rightarrow x = \frac{1}{2} y s u b s t i t u t e t o {a x}^{2} + b x + c = 0$ $a {(\frac{1}{2} y)}^{2} + b (\frac{1}{2} y) + c = 0$ $\frac{1}{4} {a y}^{2} + \frac{1}{2} b y + c = 0$ $o r \frac{1}{4} {a x}^{2} + \frac{1}{2} b x + c = 0$ ${a x}^{2} + 2 b x + 4 c = 0$
Commented by Rasheed Soomro last updated on 24/Aug/16

$N e w i d e a f o r m e! T h e r e w a s a m i s t a k e$ $i n m y a n s w e r . I h a v e c o r r e c t e d n o w .$ $B y^{'} s y m e t r i c {r o o t s}^{'} y o u m e a n p r o p o r t i o n a l$ $r o o t s ? s u c h a s k α a n d k β ?$
Commented by sandy_suhendra last updated on 25/Aug/16

$n o t j u s t s u c h a s k α a n d k β, b u t a l s o l i k e (k α + c) a n d (k β + c) o r α^{2} a n d β^{2}$
Commented by Rasheed Soomro last updated on 25/Aug/16

$TH a n k S!$
	Answered by Rasheed Soomro last updated on 24/Aug/16

	$I f α a n d β a r e t h e r o o t s o f {a x}^{2} + b x + c = 0,$ $t h e n α + β = - \frac{b}{a} a n d α β = \frac{c}{a}$ $F o r m u l a f o r d e t e r m i n i n g e q u a t i o n, w h o s e$ $^{'} s u m o f t h e {r o o t s}^{'} a n d^{'} p r o d u c t o f t h e {r o o t s}^{'} a r e g i v e n .$ $x^{2} - (s u m o f t h e r o o t s) x + (p r o d u c t o f t h e r o o t s) = 0$ $(i) 2 α, 2 β$ $R e q u i r e d e q u a t i o n h a s$ $s u m o f t h e r o o t s = 2 α + 2 β = 2 (α + β) = 2 (- \frac{b}{a}) = - \frac{2 b}{a}$ $p r o d u c t o f t h e r o o t s = 2 α . 2 β = 4 α β = 4 (\frac{c}{a}) = \frac{4 c}{a}$ $R e q u i r e d e q u a t i o n w i l l b e$ $x^{2} - (- \frac{2 b}{a}) x + (\frac{4 c}{a}) = 0$ ${a x}^{2} + 2 b x + 4 c = 0$ $(i i) α^{2}, β^{2}$ $S u m o f t h e r o o t s o f r e q u i r e d e q u a t i o n$ $= α^{2} + β^{2} = {(α + β)}^{2} - 2 α β$ $= {(- \frac{b}{a})}^{2} - 2 (\frac{c}{a}) = \frac{b^{2} - a c}{a}$ $P r o d u c t o f t h e r o o t s o f r e q u i r e d e q u a t i o n$ $= α^{2} . β^{2} = {(α β)}^{2} = {(\frac{c}{a})}^{2} = \frac{c^{2}}{a^{2}}$ $R e q u i r e d e q u a t i o n :$ $x^{2} - (\frac{b^{2} - a c}{a}) x + (\frac{c^{2}}{a^{2}}) = 0$ $a^{2} x^{2} - a (b^{2} - a c) x + c^{2} = 0$ $(i i i) α + 1, β + 1$ $S u m o f t h e r o o t s o f r e q u i r e d e q u a t i o n$ $= (α + 1) + (β + 1) = (α + β) + 2$ $= (- \frac{b}{a}) + 2 = \frac{2 - b}{a}$ $P r o d u c t o f t h e r o o t s o f r e q u i r e d e q u a t i o n$ $= (α + 1) . (β + 1) = α β + α + β + 1$ $= (\frac{c}{a}) + (- \frac{b}{a}) + 1 = \frac{a - b + c}{a}$ $R e q u i r e d e q u a t i o n :$ $x^{2} - (\frac{2 - b}{a}) x + (\frac{a - b + c}{a}) = 0$ ${a x}^{2} + (b - 2) x + (a - b + c) = 0$ $(i v) \frac{1 + α}{1 - α}, \frac{1 + β}{1 - β}$ $S u m o f t h e r o o t s = \frac{1 + α}{1 - α} + \frac{1 + β}{1 - β} = \frac{(1 + α) (1 - β) + (1 + β) (1 - α)}{(1 - α) (1 - β)}$ $= \frac{(1 - β + α - α β) + (1 - α + β - α β)}{1 - β - α + α β}$ $= \frac{1 + (α - β) - α β + 1 - (α - β) - α β}{1 - (α + β) + α β} = \frac{2 - 2 α β}{1 - (α + β) + α β} = \frac{2 - 2 (\frac{c}{a})}{1 - (- \frac{b}{a}) + \frac{c}{a}}$ $= \frac{\frac{2 a - 2 c}{a}}{\frac{a + b + c}{a}} = \frac{2 (a - c)}{a + b + c}$ $P r o d u c t o f t h e r o o t s = \frac{1 + α}{1 - α} \times \frac{1 + β}{1 - β} = \frac{1 + (α + β) + α β}{1 - (α + β) + α β} = \frac{1 + (- \frac{b}{a}) + (\frac{c}{a})}{1 - (- \frac{b}{a}) + (\frac{c}{a})}$ $= \frac{a - b + c}{a + b + c}$ $R e q u i r e d e q u a t i o n : x^{2} - (\frac{2 (a - c)}{a + b + c}) x + \frac{a - b + c}{a + b + c} = 0$ $(a + b + c) x^{2} - 2 (a - c) x + (a - b + c) = 0$ $(v) \frac{α}{β}, \frac{β}{α}$ $(v) S u m o f t h e r o o t s = \frac{α}{β} + \frac{β}{α} = \frac{α^{2} + β^{2}}{α β} = \frac{{(α + β)}^{2} - 2 α β}{α β}$ $= \frac{{(- \frac{b}{a})}^{2} - 2 (\frac{c}{a})}{\frac{c}{a}} = \frac{b^{2} - 2 a c}{a^{2}} \times \frac{a}{c} = \frac{b^{2} - 2 a c}{a c}$ $P r o d u c t o f t h e r o o t s = \frac{α}{β} \times \frac{β}{α} = 1$ $r e q u i r e d e q u a t i o n$ $x^{2} - (\frac{b^{2} - 2 a c}{a c}) x + 1 = 0$ ${a c x}^{2} - (b^{2} - 2 a c) x + a c = 0$
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