First slide

Theory of equations

Question

The roots of the given equation $(p - q) x^{2} + (q - r) x + (r - p) = 0$ are

Easy

A

$\frac{p - q}{r - p}, 1$
B

$\frac{q - r}{p - q}, 1$
C

$\frac{r - p}{p - q}, 1$
D

$1, \frac{q - r}{p - q}$

Solution

Given equation is
$\begin{array}{l} (p - q) x^{2} + (q - r) x + (r - p) = 0 \\ \Rightarrow x = \frac{(r - q) \pm \sqrt{(q - r)^{2} - 4 (r - p) (p - q)}}{2 (p - q)} \\ = \frac{(r - q) \pm \sqrt{q^{2} + r^{2} - 2 qr - 4 (rp - rq - p^{2} + pq)}}{2 (p - q)} \\ \Rightarrow x = \frac{(r - q) \pm (q + r - 2 p)}{2 (p - q)} \\ \Rightarrow x = \frac{r - p}{p - q}, 1 \end{array}$

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