# How do you solve  x = sqrtx + 6?

Jul 20, 2015

Substitute $t = \sqrt{x}$ to get a quadratic in $t$. Solve for $t$, then square to find $x = 9$

#### Explanation:

Let $t = \sqrt{x}$. Note $t \ge 0$.

Then ${t}^{2} = t + 6$

Subtract $t + 6$ from both sides to get:

$0 = {t}^{2} - t - 6 = \left(t - 3\right) \left(t + 2\right)$

So $t = 3$ or $t = - 2$.

Discard $t = - 2$ since $t$ is the non-negative square root of $x$.

So $t = 3$ and $x = {t}^{2} = {3}^{2} = 9$