treetopssan

2022-01-30

Why do I get incorrect solutions $x=0$ and $x=2$ for $x+1+\sqrt{4x+1}=0?$

basgrwthej

Step 1
Here is one correct way to solve this problem I learnt from comments.
I should write
$x+1=-\sqrt{4x+1}$
$\phantom{\rule{0.278em}{0ex}}⟺\phantom{\rule{0.278em}{0ex}}\left\{\begin{array}{l}\left(x+1{\right)}^{2}=4x+1\\ \left(x+1\right)\le 0\end{array}$
$\phantom{\rule{0.278em}{0ex}}⟺\phantom{\rule{0.278em}{0ex}}\left\{\begin{array}{l}x\left(x-2\right)=0\\ x\le -1\end{array}$
Then I should solve $x\left(x-2\right)=0$ to get $x=0$ and $x=2$. Then I should reject both solutions because both solutions do not satisfy $x\le 1$. Therefore the equation has no solutions.

Karly Logan

Step 1
You can easily track where the extra solutions were introduced
$x+1+\sqrt{4x+1}=0\to 0+1+\sqrt{1}=0$
$x+1=-\sqrt{4x+1}\to 0+1=-\sqrt{0+1}$
${x}^{2}+2x+1=4x+1\to 0+0+1=0+1$
${x}^{2}-2x=0\to 0-0=0$
$x\left(x-2\right)=0\to 0\left(0-2\right)=0$
Repeat with $x=2$.
If you turn $a=b$ to ${a}^{2}={b}^{2}$, you also allow $a=-b$.

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