The problem asks us to find the range of values for the constant $a$ such that the equation $4^x - a \cdot 2^{x+1} + a^2 + a - 6 = 0$ has two distinct real solutions for $x$.

1. Problem Description

The problem asks us to find the range of values for the constant $a$ such that the equation $4^x - a \cdot 2^{x+1} + a^2 + a - 6 = 0$ has two distinct real solutions for $x$.

2. Solution Steps

First, rewrite the given equation:

$4^x - a \cdot 2^{x+1} + a^2 + a - 6 = 0$

$(2^x)^2 - 2a \cdot 2^x + a^2 + a - 6 = 0$

Let $t = 2^x$. Since $x$ is a real number, $t > 0$.

Then the equation becomes

$t^2 - 2at + a^2 + a - 6 = 0$

For the original equation to have two distinct real solutions for $x$, the quadratic equation in $t$ must have two distinct positive solutions for $t$.

Let $f(t) = t^2 - 2at + a^2 + a - 6$.

For $f(t) = 0$ to have two distinct positive solutions, we need the following conditions to be satisfied:

1. The discriminant must be positive: $D > 0$

2. The sum of the roots must be positive: $t_1 + t_2 > 0$

3. The product of the roots must be positive: $t_1 t_2 > 0$

$D = (-2a)^2 - 4(1)(a^2 + a - 6) = 4a^2 - 4a^2 - 4a + 24 = -4a + 24$

$D > 0 \implies -4a + 24 > 0 \implies 4a < 24 \implies a < 6$

The sum of the roots is $t_1 + t_2 = \frac{-(-2a)}{1} = 2a$.

$2a > 0 \implies a > 0$

The product of the roots is $t_1 t_2 = \frac{a^2 + a - 6}{1} = a^2 + a - 6$.

$a^2 + a - 6 > 0 \implies (a+3)(a-2) > 0$

This implies $a < -3$ or $a > 2$.

Combining these conditions:

1. $a < 6$

2. $a > 0$

3. $a < -3$ or $a > 2$

Since $a > 0$, the condition $a < -3$ is not possible.

Therefore, we need $a > 0$, $a < 6$, and $a > 2$.

So the solution is $2 < a < 6$.

3. Final Answer

$2 < a < 6$