Question

If the product of the roots of the equation $x^2-3kx+2e^{2logk}-1=0$ is $7$, then the roots of the equation are real if $k$ equals-

• A. $1$
• B. $3$
• C. $-4$
• D. $\pm 2$

Answer 1

The correct option is D $\pm 2$

Given

$x^2-3kx+2e^{2logk}-1=0$

$\Rightarrow x^2-3kx+2e^{log{e}^{k^2}}-1=0$

$\Rightarrow x^2-3kx+2k^2-1=0$

$\alpha \beta =7$

$\Rightarrow \frac{2k^2-1}{1}=7$

$\Rightarrow 2k^2=8$

$\Rightarrow k=\pm 2$

$\therefore k=\pm 2$

But $logk$ is defined for $k>0$

$\therefore k=2$