First slide

Theory of equations

Question

The product of the roots of the equation $x^{2} - 4 m x + 3 e^{2 \log m} - 4 = 0$ is 8, then its roots will be real when $m$ equals

Moderate

A

1
B

$\sqrt{2}$
C

$\pm 2$
D

$\pm \sqrt{2}$

Solution

The given equation is $x^{2} - 4 m x + 3 e^{2 \log m} - 4 = 0$
Given that, product of the roots $= 8$
$\Rightarrow$    $3 e^{2 \log m} - 4 = 8$
$\Rightarrow$    $3 e^{\log m^{2}} = 12$    $\Rightarrow$    $m^{2} = 4$
$\Rightarrow$    $m = \pm 2$ ..
Now the given equation becomes $x^{2} - 4 m x + 3 m^{2} - 4 = 0$
Its roots are real if $16 m^{2} - 4 (3 m^{2} - 4) \geq 0$    $(∵ b^{2} - 4 a c \geq 0)$
If $4 m^{2} + 16 \geq 0$ , which is true if $m = \pm 2$ .
Hence $m = \pm 2$ .

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