find the real solutions of the equation.\nx = 2\\sqrt{2x - 4}\nwhat is the solution set? select the correct…

Explanation:

Step1: Square both sides

Square both sides of the equation $x = 2\sqrt{2x - 4}$ to get $x^{2}=(2\sqrt{2x - 4})^{2}$. Since $(ab)^n=a^nb^n$, we have $x^{2}=4(2x - 4)$.

Step2: Expand the right - hand side

Expand $4(2x - 4)$ to get $x^{2}=8x - 16$.

Step3: Rearrange to quadratic form

Rearrange the equation to the standard quadratic form $ax^{2}+bx + c = 0$. So, $x^{2}-8x + 16 = 0$.

Step4: Factor the quadratic equation

The quadratic equation $x^{2}-8x + 16$ can be factored as $(x - 4)^{2}=0$ because $(a - b)^2=a^{2}-2ab + b^{2}$ and here $a=x$ and $b = 4$.

Step5: Solve for x

Taking the square root of both sides of $(x - 4)^{2}=0$, we get $x-4 = 0$, so $x = 4$.

Step6: Check the solution

Substitute $x = 4$ into the original equation. The left - hand side is $x=4$, and the right - hand side is $2\sqrt{2\times4-4}=2\sqrt{8 - 4}=2\sqrt{4}=2\times2 = 4$. Since the left - hand side equals the right - hand side when $x = 4$, $x = 4$ is a valid solution.

Answer:

A. 4