Determine real number exists for relation with square roots

Question

We have $$\sqrt{x -2} = 3 -2\sqrt{x}$$.

I am to find whether a real number exists for this relation, and the real number that satisfies.

I start by squaring both sides, which yields:

$$x - 2 = 4x - 12\sqrt{x} + 9$$.

Whence:

$$ -3x = -12\sqrt{x} + 11 \\ \sqrt{x} = \frac{x}{4} + \frac{11}{12}. $$

But once i get here i am stuck. How can i find whether a solution exists for x from here?

Answer 1

Again: square both sides and solve the resulting quadratic. You'll get

$$x = \left( \frac{x}{4} + \frac{11}{12}\right)^{\! 2}$$

Remember to check your answers in the original question.

Answer 2

Let $y=\sqrt{x}$. Then, $$ \sqrt{x} = \frac{x}{4} + \frac{11}{12}\\ \implies 0 = \frac{y^2}{4} -y + \frac{11}{12} $$ You should be able to use the quadratic formula to solve for $y$ (check for extraneous solutions), and then don't forget to solve for $x$ from $y$.