If $\frac{x^2-bx}{ax-c} = \frac{k-1}{k+1}$ has roots, whose magnitude is equal but signs are opposite.

Question

If $\frac{x^2-bx}{ax-c} = \frac{k-1}{k+1}$ has roots, whose magnitude is equal but signs are opposite.

Answer is $\frac{a-b}{a+b}$

I used cross multiplication and since the roots are opposite in sign, on adding the roots, the total must be zero. But this is a long method. Please tell me any shorter method to solve this problem.

lab bhattacharjee · Accepted Answer · 2019-04-26T08:49:56.880

1

Assuming we need the value of $k,$

$$(k+1)x^2-x[b(k+1)-a(k-1)]+c(k-1)=0$$

So, if $\alpha$ is a root, $-\alpha$ will be the other

$$\implies \alpha+(-\alpha)=\dfrac{b(k+1)-a(k-1)}{k+1}$$

$$\implies\dfrac{k-1}{k+1}= \dfrac ab$$

Apply Componendo et Dividendo

edited Apr 26 '19 at 08:49

answered Apr 26 '19 at 08:31

lab bhattacharjee

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I am getting $ \frac{k+1}{k-1} x^2 -x(b\frac{k+1}{k-1} -a) -c =0 $ – sawan kumawat Apr 26 '19 at 08:42
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@user168282, Please find the updated answer – lab bhattacharjee Apr 26 '19 at 08:50

If $\frac{x^2-bx}{ax-c} = \frac{k-1}{k+1}$ has roots, whose magnitude is equal but signs are opposite.

1 Answers1