The values of k for which the quadratic equation

(c) 5, −7

The given equation is $k x^{2}+1=k x+3 x-11 x^{2}$ which can be written as.

$k x^{2}+11 x^{2}-k x-3 x+1=$

$\Rightarrow(k+11) x^{2}-(k+3) x+1=0$

For equal and real roots, the discriminant of $(k+11) x^{2}-(k+3) x+1=0$.

$\therefore(k+3)^{2}-4(k+11)=0$

$\Rightarrow k^{2}+2 k-35=0$

$\Rightarrow(k-5)(k+7)=0$

$\Rightarrow k=5,-7$

Hence, the equation has real and equal roots when $k=5,-7$.