Solution to this radical equation?

Question

Is $-1$ a solution to $\sqrt{4x+5}=x$ ?
Why or why not?

Answer 1

If you are looking for real solutions, there is one. Indeed,

\begin{align*} \sqrt{4x+5} = x \Longleftrightarrow \begin{cases} 4x + 5 = x^{2}\\ x \geq 0 \end{cases} \Longleftrightarrow \begin{cases} x^{2} - 4x - 5 = 0\\ x \geq 0 \end{cases} \Longleftrightarrow \begin{cases} (x-2)^{2} - 9 = 0\\ x\geq 0 \end{cases} \end{align*}

Therefore the solution set is $S = \{5\}$, since the other root is negative.

Answer 2

$$\sqrt{4x+5}=x$$

$=(\sqrt{4x+5})^2=x^2$

$=4x+5=x^2$

$0=x^2-4x-5$

$0=(x-5)(x+1)$

So $x=5$ is a solution (real). $x=-1$ is not solution since our range is greater than zero. By squaring both sides, we made negative solutions possible so we had to reject them at the end.