The sum of the solutions of the equation $$\left|\sqrt{x}-2\left|+\sqrt{x}\left(\sqrt{x}-4\right)+2=0\right|\right|$$, $$x > 0$$ is equal to:

To solve this equation cleanly, let's substitute $$t = \sqrt{x}$$. Since $$x > 0$$, we must have $$t > 0$$.

Rewriting the equation in terms of $$t$$:

$$|t - 2| + t(t - 4) + 2 = 0$$  $$|t - 2| + t^2 - 4t + 2 = 0$$

Now, we handle the absolute value by splitting the problem into two cases based on the critical point $$t = 2$$.

Case 1: When $$t \geq 2$$

For $$t \geq 2$$, the expression inside the absolute value is non-negative, so $$|t - 2| = t - 2$$.

Substitute this back into the equation:

$$(t - 2) + t^2 - 4t + 2 = 0$$  $$t^2 - 3t = 0$$  $$t(t - 3) = 0$$

This gives two potential roots: $$t = 0$$ or $$t = 3$$.

• Since our condition for this case is $$t \geq 2$$, we reject $$t = 0$$ and accept $$t = 3$$.

Case 2: When $$0 < t < 2$$

For $$t < 2$$, the expression inside the absolute value is negative, so $$|t - 2| = -(t - 2) = 2 - t$$.

Substitute this back into the equation:

$$(2 - t) + t^2 - 4t + 2 = 0$$  $$t^2 - 5t + 4 = 0$$

Factor the quadratic equation:

$$(t - 1)(t - 4) = 0$$

This gives two potential roots: $$t = 1$$ or $$t = 4$$.

• Since our condition for this case is $$0 < t < 2$$, we reject $$t = 4$$ and accept $$t = 1$$.

Step 3: Find the values of $$x$$ and their sum

From our accepted values of $$t$$, we solve for $$x$$ using $$x = t^2$$:

1. For $$t = 3 \implies x = 3^2 = 9$$
2. For $$t = 1 \implies x = 1^2 = 1$$

Both solutions ($$x = 1$$ and $$x = 9$$) satisfy the original condition $$x > 0$$.

The sum of the solutions is:

$$\text{Sum} = 9 + 1 = 10$$

Answer:

10