First slide

Theory of equations

Question

If one root of the equation $a x^{2} + b x + c = 0$ is reciprocal of the one root of the equation $a_{1} x^{2} + b_{1} x + c_{1} = 0$ , then

Moderate

A

${(a a_{1} - c c_{1})}^{2} = (b c_{1} - b_{1} a) (b_{1} c - a_{1} b)$
B

${(a b_{1} - a_{1} b)}^{2} = (b c_{1} - b_{1} c) (c a_{1} - c_{1} a)$
C

${(b c_{1} - b_{1} c)}^{2} = (c a_{1} - a_{1} c) (a b_{1} - a_{1} b)$
D

none of these

Solution

Let $a$ be a root of the equation $a x^{2} + b x + c = 0$
Then , $1 / α$ is a root of $a_{1} x^{2} + b_{1} x + c_{1} = 0$
$\Rightarrow a α^{2} + b α + c = 0$ (i)
and , $\frac{a_{1}}{α^{2}} + \frac{b_{1}}{α} + c_{1} = 0 \Rightarrow c_{1} α^{2} + b_{1} α + a_{1} = 0$ (ii)
From (i) and (ii), we have
$\begin{array}{l} \frac{α^{2}}{b a_{1} - b_{1} c} = \frac{α}{c c_{1} - a a_{1}} = \frac{1}{a b_{1} - c_{1} b} \\ \Rightarrow a^{2} = \frac{b a_{1} - b_{1} c}{a b_{1} - c_{1} b}, α = \frac{c c_{1} - a a_{1}}{a b_{1} - c_{1} b} \end{array}$
Now , $α^{2} = (α)^{2} \Rightarrow (b a_{1} - b_{1} c) (a b_{1} - c_{1} b) = {(c c_{1} - a a_{1})}^{2}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App