Consider the equation $(\log_2 x)^2-4\log_2x-(m^2+2m+13)=0$. Let the real roots of the equation be $x_1,x_2$ such that $x_1<x_2$. Find the sum of maximum value of $x_1$ and minimum value of $x_2$
My attempt is as follows:-
For minimizing $x_2-x_1$, we need to minimize $x_2$ and maximize $x_1$ and I think vice-versa should also be true.
Let $u=x_2-x_1$ and assume $y=\log_2 x$, so equation would look like $y^2-4y-(m^2+2m+13)=0$
$$y_1+y_2=4$$ $$y_1y_2=-(m^2+2m+13)$$
$$u=x_2-x_1=2^{y_2}-2^{y_1}$$
$$u=2^{y_2}-2^{4-y_2}$$
$$\dfrac{du}{d{y_2}}=2^{y_2}\log 2+2^{4-y_2}\log2$$
$$2^{y_2}+\dfrac{16}{2^{y_2}}=0$$
But its not taking me anywhere.
Any suggestions or inputs? Actual answer is $\dfrac{257}{4}$
$$x_2=2^{2+\sqrt{m^2+2m+17}}=2^{2+\sqrt{(m+1)^2+16}}\geq2^{2+4}=64.$$ $$x_1=2^{2-\sqrt{m^2+2m+17}}=2^{2-\sqrt{(m+1)^2+16}}\leq2^{2-4}=\frac{1}{4}.$$ The equality in the both cases occurs for $m=-1$, which gives the answer: $64\frac{1}{4}.$