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I need to find the solutions of $$x=\sqrt{5+4x}.$$ I found that $x=5 \vee x=-1$. But is $-1=\sqrt{1}$ really true?

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hum,how did you find that?I would like to be enlightened –  aflous Apr 30 at 10:56
$x=\sqrt{5+4x} \implies x^2 = 5+4x \implies x^2-4x-1=0 \implies (x-5)(x+1)=0$ –  Graham Kemp Apr 30 at 11:00
Having found this you must always check (as you did). The right conclusion is that $x=5$. It appears to satisfy. This not the case for $x=-1$. This because $1=\sqrt{1}$ so $-1\ne\sqrt{1}$ –  drhab Apr 30 at 11:04
@aflous At first I squared both sides. Subsequently I completed the square: $(x-2)^2=9$. But because squaring isn't an equivalent transformation I checked both solutions. –  Richard Apr 30 at 11:04
That's what I wanted you to say:squaring isn't an equivalent transformation, so you got your answer –  aflous Apr 30 at 11:11

2 Answers 2

up vote 7 down vote accepted

Well, the only candidates possible for the solution are $x=-1$ and $x=5$. However, since in the process of getting these candidates, you had to square the equation, it may happen that you got some solutions that are not true, so you must inspect your candidates.

The first, $x=5$, put in the original equation, yields $5=\sqrt{25}$ which is true.

The second, however, yields $-1=\sqrt1$ which is not true, because $\sqrt{1} = 1$ (note that $\sqrt{a}$ is defined as the positive number such that its square is $a$).

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Thanks! The information in your last sentence after "(" is exactly what I needed to know. –  Richard Apr 30 at 11:18
Happy to help. Just a quick fun fact: the direct consequence of my last sentence is the fact that $\sqrt{x^2}$ is actually not $x$, but $|x|$. –  5xum Apr 30 at 11:20

But if u check this link:


it shows that principal root of $\sqrt(1)$ is 1. But -1 is also a real root of $sqrt(1)$

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