First slide

Theory of expressions

Question

If the sum of the roots of the equation $a x^{2} + b x + c = 0$ is equal to the sum of the squares of their reciprocals, then

Moderate

A

$c^{2} b, a^{2} c, b^{2} a$ are in A.P.
B

$c^{2} b, a^{2} c, b^{2} a$ are in G. P.
C

$\frac{b}{c}, \frac{a}{b}, \frac{c}{a}$ are in G.P.
D

$\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$ are in G.P.

Solution

Let $α, β$ be the roots of the given quadratic equation. Then,
$α + β = - b / a, α β = c / a$
It is given that
$\begin{array}{l} α + β = \frac{1}{α^{2}} + \frac{1}{b^{2}} \\ \Rightarrow α^{2} + β^{2} = (α + β) α^{2} β^{2} \\ \Rightarrow (α + β)^{2} - 2 α β = (α + β) (α β)^{2} \\ \Rightarrow \frac{b^{2}}{a^{2}} - \frac{2 c}{a} = \frac{- b c^{2}}{a^{3}} \end{array}$
$\Rightarrow \frac{2 c}{a} = \frac{b^{2}}{a^{2}} + \frac{b c^{2}}{a^{3}}$
$\Rightarrow 2 a^{2} c = a b^{2} + b c^{2} \Rightarrow c^{2} b, a^{2} c, b^{2} a$ are in A.P.
Dividing both sides of $2 a^{2} c = a b^{2} + b c^{2}$ by $a b c$ , we get
$2 \frac{a}{b} = \frac{b}{c} + \frac{c}{a} \Rightarrow \frac{b}{c}, \frac{a}{b}, \frac{c}{a}$ are in A.P.

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