First slide

Theory of equations

Question

If the roots of the equation ${px}^{2} + 2 qx + r = 0$ and ${qx}^{2} - 2 \sqrt{pr} x + q = 0$ be real, then

Moderate

A

$p = q$
B

$q^{2} = pr$
C

$p^{2} = qr$
D

$r^{2} = pq$

Solution

Equations ${px}^{2} + 2 qx + r = 0$ and ${qx}^{2} - 2 (\sqrt{pr}) x + q = 0$ have real roots, then from first
$4 q^{2} - 4 pr \geq 0 \Rightarrow q^{2} - pr \geq 0 \Rightarrow q^{2} \geq pr . . . [i] and from second 4 (pr) - 4 q^{2} \geq 0 (for real root) \Rightarrow pr \geq q^{2} . . . [ii] From [i] and [ii], we get result q^{2} = pr .$

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