Question

If the product of the roots of the equation $x^2-2\sqrt{2}kx+2e^{2\log \left(k\right)}-1=0$ is $31$, then the roots of the equation are real for $k$, is equal to

• A. $-4$
• B. $1$
• C. $4$
• D. $0$

Answer 1

The correct option is C

$4$

Explanation of the correct option.

It is given that the product of the roots of the equation $x^2-2\sqrt{2}kx+2e^{2\log \left(k\right)}-1=0$ is $31$,

We know that for $a{x}^2+bx+c=0$ product of roots is given by $\frac{c}{a}$.

Thus, $2e^{2\log \left(k\right)}-1=31$

$\Rightarrow$ $2e^{2\log \left(k\right)}=32$

$\Rightarrow$ $k^2=16$ $\left[\because e^{a\log \left(x\right)}=x^a\right]$

$\Rightarrow$ $k=\pm 4$

Since $\log \left(k\right)$ is defined for only a positive value of $k$, thus, $k\ne -4$.

Therefore, $k=4$.

Hence option $\left(C\right)$ is the correct option.