First slide

Theory of equations

Question

If both the roots of $k (6 x^{2} + 3) + rx + 2 x^{2} - 1 = 0$ and $6 k (2 x^{2} + 1) + px + 4 x^{2} - 2 = 0$ are common, then $2 r - p$ is equal to

Moderate

A

$- 1$
B

$0$
C

$1$
D

$2$

Solution

Given equation can be written as
$(6 k + 2) x^{2} + rx + 3 k - 1 = 0 . . . (i)$
and $2 (6 k + 2) x^{2} + px + 2 (3 k - 1) = 0 . . . (ii)$
Condition for common roots is
$\frac{12 k + 4}{6 k + 2} = \frac{p}{r} = \frac{6 k - 2}{3 k - 1} = 2 or 2 r - p = 0$

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