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

P97
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    No, because $\sqrt{4x+5} \geq 0$. – user0102 Sep 17 '19 at 20:04
  • Sure it is, but make sure to mention that you don't restrict yourself to the principal branch of the square root or a lot of people will get mad. – Hyperplane Sep 17 '19 at 20:09
  • At the same time, it teaches you that, when you square both sides of the equation (obtaining $4x+5=x^2$), you don't end up with an equivalent equation. What you get may have solutions that the original equation did not have. Precisely, you get $-1$ and $5$, but only $5$ is the solution of the original equation. –  Sep 17 '19 at 20:09

2 Answers2

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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.

user0102
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$$\sqrt{4x+5}=x$$

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

$=4x+5=x^2$

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

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

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.

Sina Babaei Zadeh
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