First slide

Theory of equations

Question

If a, b, c are in H.P., then the roots of the equation $a (b - c) x^{2} + b (c - a) x + c (a - b) = 0$

Moderate

A

$1, 1$
B

$1, \frac{a (b - c)}{c (a - b)}$
C

$1, \frac{a (b - c)}{b (c - a)}$
D

None of these

Solution

We know, that a (b -c) +b (c -a) + c (a-b) =0.
So, x =1 is a root of the given equation.
Let $α$ be the other root. Then
$\begin{aligned} α \times 1 = \frac{c (a - b)}{a (b - c)} \Rightarrow α = \frac{\frac{1}{b} - \frac{1}{a}}{\frac{1}{c} - \frac{1}{b}} \\ \Rightarrow & α = 1 [∵ a, b, c are in H.P. ∴ \frac{1}{a}, \frac{1}{b}, \frac{1}{c} are in A.P.] \end{aligned}$
Thus, the given equation has equal roots each equal to 1.

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