11.17 For what values of $p$ does the equation

$x^{2}-\left(2^{p}-1\right) x-3\left(4^{p-1}-2^{p-2}\right)=0$

have equal roots?
11.17 A quadratic equation has equal roots if its discriminant $D=b^{2}-4 a c$ is equal to zero. We find

$$
\begin{aligned}
& D=\left(2^{p}-1\right)^{2}+12\left(4^{p-1}-2^{p-2}\right)= \\
& =2^{2 p}-2 \cdot 2^{p}+1+\frac{12 \cdot 2^{2 p}}{4}-\frac{12 \cdot 2^{p}}{4}=4 \cdot 2^{2 p}-5 \cdot 2^{p}+1
\end{aligned}
$$

Using the substitution $2^{p}=y$, we get the equation $4 y^{2}-5 y+1=0$, which has roots $y_{1}=\frac{1}{4}$, $y_{2}=1$. Solving the equations $2^{p}=2^{-2}$ and $2^{p}=1$, we get $p=-2$ and $p=0$.

Answer: for $p=-2$ and $p=0$.