# How do you solve sqrt(9x-8)=x?

Aug 10, 2015

$\left\{\begin{matrix}x = 8 \\ x = 1\end{matrix}\right.$

#### Explanation:

Right from the start, you know that your solution(s) must satisfy two conditions

• $x \ge 0$
• $9 x - 8 \ge 0 \implies x \ge \frac{8}{9}$

Overall, you need your solution(s) to satisfy $x \ge \frac{8}{9}$.

Start by squaring both sides of the equation to get rid of the radical term

${\left(\sqrt{9 x - 8}\right)}^{2} = {x}^{2}$

$9 x - 8 = {x}^{2}$

This is equivalent to

${x}^{2} - 9 x + 8 = 0$

The two solutions to this quadratic can be found using the quadratic formula

${x}_{1 , 2} = \frac{- \left(- 9\right) \pm \sqrt{{\left(- 9\right)}^{2} - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{9 \pm \sqrt{49}}{2}$

${x}_{1 , 2} = \frac{9 \pm 7}{2} = \left\{\begin{matrix}{x}_{1} = \frac{9 + 7}{2} = 8 \\ {x}_{2} = \frac{9 - 7}{2} = 1\end{matrix}\right.$

Since both ${x}_{1}$ and ${x}_{2}$ satisfy the condition $x \ge \frac{8}{9}$, your original equation will have two solutions

$\left\{\begin{matrix}{x}_{1} = \textcolor{g r e e n}{8} \\ {x}_{2} = \textcolor{g r e e n}{1}\end{matrix}\right.$

Do a quick check to make sure that the calculations are correct

$\sqrt{9 \cdot 8 - 8} = 8$

$\sqrt{64} = 8 \iff 8 = 8 \textcolor{g r e e n}{\sqrt{}}$

and

$\sqrt{9 \cdot 1 - 8} = 1$

$\sqrt{1} = 1 \iff 1 = 1 \textcolor{g r e e n}{\sqrt{}}$