If the ratio of the roots of equation ax^2 +2bx+c=0 is same as the ratio of the roots of px^2 +2qx+r=0 where a,b,c,p ,r are non zero real numbers . Then the value of ((b^2 /q^2 ))((p/a))((r/c)) is equal to


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Question Number 162165 by cortano last updated on 27/Dec/21

$If the ratio of the roots of equation$ ${a x}^{2} + 2 b x + c = 0 is same as the ratio$ $of the roots of {p x}^{2} + 2 q x + r = 0 where$ $a, b, c, p, r are non zero real numbers$ $. Then the value of (\frac{b^{2}}{q^{2}}) (\frac{p}{a}) (\frac{r}{c}) is equal to$
Commented by bobhans last updated on 27/Dec/21
${ ((ax^2 +2bx+c=0⇒(x_1 /x_2 )=λ)),((px^2 +2qx+r=0⇒(x_3 /x_4 )=λ)) :}⇒(x_1 /x_2 ) = (x_3 /x_4 ) ⇒(x_2 /x_4 )=(x_1 /x_3 ) { ((x_1 .x_2 = (c/a)⇒λx_2 ^2 = (c/a) ; x_1 +x_2 =−((2b)/a))),((x_3 .x_4 =(r/p)⇒λx_4 ^2 = (r/p) ; x_3 +x_4 =−((2q)/p))) :} { ((x_2 (1+λ)=−((2b)/a)⇒1+λ=−((2b)/(ax_2 )))),((x_4 (1+λ)=−((2q)/p) ⇒1+λ=−((2q)/(px_4 )))) :} ⇒((2b)/(ax_2 )) = ((2q)/(px_4 )) ; (x_2 /x_4 ) = ((bp)/(aq)) ⇒(x_1 /x_3 ) .(x_2 /x_4 ) = ((cp)/(ar)) ⇒(c/r)=((ax_1 x_2 )/(px_3 x_4 )) ∴ ((b^2 /q^2 ))((p/a))((r/c))= (((bp)/(aq)))((b/q))((1/(c/r))) = ((x_2 /x_4 ))(((ax_2 )/(px_4 )))(((px_3 x_4 )/(ax_1 x_2 )))= ((x_2 x_3 )/(x_1 x_4 )) = 1$
${\begin{cases} {ax}^{2} + 2 bx + c = 0 \Rightarrow \frac{x_{1}}{x_{2}} = λ \\ {px}^{2} + 2 qx + r = 0 \Rightarrow \frac{x_{3}}{x_{4}} = λ \end{cases} \Rightarrow \frac{x_{1}}{x_{2}} = \frac{x_{3}}{x_{4}} \Rightarrow \frac{x_{2}}{x_{4}} = \frac{x_{1}}{x_{3}}$ ${\begin{cases} x_{1} . x_{2} = \frac{c}{a} \Rightarrow λ x_{2}^{2} = \frac{c}{a}; x_{1} + x_{2} = - \frac{2 b}{a} \\ x_{3} . x_{4} = \frac{r}{p} \Rightarrow λ x_{4}^{2} = \frac{r}{p}; x_{3} + x_{4} = - \frac{2 q}{p} \end{cases}$ ${\begin{cases} x_{2} (1 + λ) = - \frac{2 b}{a} \Rightarrow 1 + λ = - \frac{2 b}{{ax}_{2}} \\ x_{4} (1 + λ) = - \frac{2 q}{p} \Rightarrow 1 + λ = - \frac{2 q}{{px}_{4}} \end{cases}$ $\Rightarrow \frac{2 b}{{ax}_{2}} = \frac{2 q}{{px}_{4}}; \frac{x_{2}}{x_{4}} = \frac{bp}{aq}$ $\Rightarrow \frac{x_{1}}{x_{3}} . \frac{x_{2}}{x_{4}} = \frac{cp}{ar} \Rightarrow \frac{c}{r} = \frac{{ax}_{1} x_{2}}{{px}_{3} x_{4}}$ $∴ (\frac{b^{2}}{q^{2}}) (\frac{p}{a}) (\frac{r}{c}) = (\frac{bp}{aq}) (\frac{b}{q}) (\frac{1}{c / r})$ $= (\frac{x_{2}}{x_{4}}) (\frac{{ax}_{2}}{{px}_{4}}) (\frac{{px}_{3} x_{4}}{{ax}_{1} x_{2}}) = \frac{x_{2} x_{3}}{x_{1} x_{4}} = 1$
	Answered by Rasheed.Sindhi last updated on 27/Dec/21

	$L e t t h e r o o t s o f {a x}^{2} + 2 b x + c = 0$ $a r e α m α .$ $t h e r o o t s o f {p x}^{2} + 2 q x + r = 0 a r e$ $β, m β$ $∙ α + m α = - \frac{2 b}{a}, m α^{2} = \frac{c}{a}$ $α = - \frac{2 b}{(1 + m) a}, m {(- \frac{2 b}{(1 + m) a})}^{2} = \frac{c}{a}$ $∙ β + m β = - \frac{2 q}{p}, m β^{2} = \frac{r}{p}$ $β = - \frac{2 q}{(1 + m) p}, m {(- \frac{2 q}{(1 + m) p})}^{2} = \frac{r}{p}$ $\frac{m {(- \frac{2 b}{(1 + m) a})}^{2}}{m {(- \frac{2 q}{(1 + m) p})}^{2}} = \frac{\frac{c}{a}}{\frac{r}{p}}$ $\frac{\frac{4 b^{2}}{{(1 + m)}^{2} a^{2}}}{\frac{4 q^{2}}{{(1 + m)}^{2} p^{2}}} = \frac{c p}{a r}$ $\frac{4 b^{2}}{a^{2}} \times \frac{p^{2}}{4 q^{2}} = \frac{c p}{a r}$ $\frac{b^{2} p}{{a q}^{2}} = \frac{c}{r} \Rightarrow (\frac{b^{2}}{q^{2}}) (\frac{p}{a}) (\frac{r}{c}) = \frac{c}{r} (\frac{r}{c}) = 1$ $(\frac{b^{2}}{q^{2}}) (\frac{p}{a}) (\frac{r}{c}) = 1$
	Answered by Ar Brandon last updated on 27/Dec/21

	$- \frac{2 b}{a} = - \frac{2 q}{p} \Rightarrow \frac{b}{a} = \frac{q}{p} \Rightarrow \frac{b}{q} = \frac{a}{p} = \frac{c}{r}$ $\frac{c}{a} = \frac{r}{p} \Rightarrow \frac{p}{a} = \frac{r}{c}$ $(\frac{b^{2}}{q^{2}}) (\frac{p}{a}) (\frac{r}{c}) = (\frac{c^{2}}{r^{2}}) (\frac{r}{c}) (\frac{r}{c}) = 1$
	Commented by Rasheed.Sindhi last updated on 27/Dec/21

	$- \frac{2 b}{a} = - \frac{2 q}{p} ?$ $- \frac{2 b}{a} i s s u m o f t h e r o o t s i n e q n 1$ $- \frac{2 q}{p} i s s u m o f t h e r o o t s i n e q n 2$ $H o w t h e s e s u m s a r e e q u a l ?$
	Commented by Ar Brandon last updated on 28/Dec/21
	Thank you, Sir �� But I think my friend Dwaipayan Shikari was the most outstanding amongst we, the young members. I'm missing him already �� He's been offline for a very long time even on telegram. He really motivated me a lot.
	Commented by Ar Brandon last updated on 28/Dec/21

	$I love the way you often design your text, Sir .$ $E . g . T han K s S i r!$ 🤭
	Commented by Rasheed.Sindhi last updated on 28/Dec/21

	$α & m α r o o t s o f e q u a t i o n 1$ $β & m β r o o t s o f e q u a t i o n 2$ $α + m α = - \frac{2 b}{a}$ $β + m β = - \frac{2 q}{p}$ $- \frac{2 b}{a} = - \frac{2 q}{p} \Rightarrow α + m α = β + m β$ $\Rightarrow α (1 + m) = β (1 + m)$ $\Rightarrow α = β (N o t g i v e n)$
	Commented by Ar Brandon last updated on 28/Dec/21
	��
	Commented by Rasheed.Sindhi last updated on 28/Dec/21

	$α r β r a n d o n s i r, a m I w r o n g ?$
	Commented by Ar Brandon last updated on 28/Dec/21
	I'm not good enough ��
	Commented by Rasheed.Sindhi last updated on 28/Dec/21

	${Y o u}^{'} r e m o r e t h a n e n o u g h g o o d a t$ $y o u r e a r l y s t a g e o f y o u r l i f e! T h i s$ $I s a y o n t h e b a s i s o f y o u r s e n t p o s t s .$ $(I r e c a l l t h a t y o u a r e o n e o f s o m e$ $y o u n g e s t f o r u m - f r i e n d s . I^{'} m y o u r$ $o p p o s i t e i n a g e a s w e l l a s k n o w l e d g e .)$ $M a y G o d b r i g h t e n s y o u r f u t u r e!$
	Commented by Rasheed.Sindhi last updated on 28/Dec/21

	$N o d o u b t S h i k a r i w a s a l s o o u t s t a n d i n g$ $a n d I a l s o m i s s h i m . T h a n k y o u f o r$ $c o n v e r s a t i o n!$ $BTW h o w d i d y o u b r i n g t h i s e m o j i$ $i n a c o m m e n t w r i t t e n b y f o r u m$ $k e y b o a r d ?$
	Commented by Ar Brandon last updated on 28/Dec/21
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	Commented by Ar Brandon last updated on 28/Dec/21
	I clicked on line matrix then " insert plain text"
	Commented by Rasheed.Sindhi last updated on 28/Dec/21

	$T h a n k y o u A r B r a n d o n!$ 👍👍👍😃
	Commented by Ar Brandon last updated on 28/Dec/21

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