Question

The sum of value(s) of $k$ for which the equation $(({\log }_{5}k{)}^2+({\log }_{5}k)-2)x^2-(2^{2k}-34\cdot 2^k+64)x+(k^2+7k-60)=0$ possesses more than two roots, is

• A. $5$
• B. $7$
• C. $-10$
• D. $-12$

Answer 1

The correct option is A $5$

$\left(\right({\log }_{5}k{)}^2+{\log }_{5}k-2)x^2-(2^{2k}-34\cdot 2^k+64)x+(k^2+7k-60)=0$
For the equation to have more than $2$ roots, it should be an identity, so
$({\log }_{5}k{)}^2+{\log }_{5}k-2=0\Rightarrow ({\log }_{5}k+2\left)\right({\log }_{5}k-1)=0\Rightarrow {\log }_{5}k=-2,1\Rightarrow k=\frac{1}{25},5\cdots (1)$

$2^{2k}-34\cdot 2^k+64=0\Rightarrow (2^k-32)(2^k-2)=0\Rightarrow 2^k=2,32\Rightarrow k=1,5\cdots \left(2\right)$

$k^2+7k-60=0\Rightarrow (k+12)(k-5)=0\Rightarrow k=-12,5\cdots \left(3\right)$

From equations $\left(1\right),\left(2\right)$ and $\left(3\right),$
$k=5$
Hence, sum of all values of $k$ is $5.$