Question

For what values of $k$ does the equation $k{x}^2+12x-3=0$ possess a root equal to $\frac{1}{5}$?

Answer 1

$K{x}^2+12x-3=0$

Let a root equal to $\frac{1}{5}$.

$K-(\frac{1}{5}{)}^2+12(\frac{1}{5})-3=0$

$\frac{K^2}{25}+\frac{12}{5}-3=0$

$K^2+60-75=0$

$K^2=15$

$K=\pm \sqrt{15}$