Question

If $k>0$ and the product of roots of the equations $x^2-3kx+2e^{\log k}-1=0$ is $7$ , then find the sum of the roots-

• A. $6$
• B. $4$
• C. $3$
• D. $-12$

Answer 1

The correct option is A $6$

$\begin{array}{c}{x}^2-3kx+2.e^{2\log k}-1=0\\ x^2-3kx+2e^{\log k^2}-1=0\\ x^2-3kx+2e^{\log k^2}-1=0\\ x^2-3kx+\left(2k^2-1\right)=0\\ Letalphaand\beta betheroots\\ \alpha +\beta =3k\\ \alpha \beta =2k^2-1=7\\ 2k^2=8=4\\ k=\pm 2\\ k=2k>0\\ Now,\\ \alpha +\beta =3k=3\left(2\right)=6\\ Hence,optionaisthecorrectanswer.\end{array}$