How do you solve 4-x=sqrt(x-4)?

Oct 17, 2015

$x = 4$

Explanation:

The easy way

$\sqrt{x - 4}$ only takes Real values when $x \ge 4$ and then $\sqrt{x - 4} \ge 0$.

When $x \ge 4$ then $4 - x \le 0$

So the only solution is $x = 4$

The normal way

Square both sides of the equation to get:

$16 - 8 x + {x}^{2} = {\left(4 - x\right)}^{2} = {\left(\sqrt{x - 4}\right)}^{2} = x - 4$

Note that squaring both sides of an equation may introduce spurious solutions (as it does in this case).

Subtract $x - 4$ from both ends to get:

${x}^{2} - 9 x + 20 = 0$

This factors as:

$\left(x - 4\right) \left(x - 5\right) = 0$

So $x = 4$ or $x = 5$.

Then check the solutions:

If $x = 4$ then $4 - x = 0$ and $\sqrt{x - 4} = 0$. So $x = 4$ is a solution of the original equation.

If $x = 5$ then $4 - x = - 1$ but $\sqrt{x - 4} = 1$. So $x = 5$ is not a solution of the original equation.

Note that this happens because numbers have two square roots. When you square both sides of an equation you throw away some information.