treetopssan

2022-01-30

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

basgrwthej

Beginner2022-01-31Added 13 answers

Step 1

Here is one correct way to solve this problem I learnt from comments.

I should write

$x+1=-\sqrt{4x+1}$

$\phantom{\rule{0.278em}{0ex}}}\u27fa{\textstyle \phantom{\rule{0.278em}{0ex}}}\{\begin{array}{l}(x+1{)}^{2}=4x+1\\ (x+1)\le 0\end{array$

$\phantom{\rule{0.278em}{0ex}}}\u27fa{\textstyle \phantom{\rule{0.278em}{0ex}}}\{\begin{array}{l}x(x-2)=0\\ x\le -1\end{array$

Then I should solve$x(x-2)=0$ 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.

Here is one correct way to solve this problem I learnt from comments.

I should write

Then I should solve

Karly Logan

Beginner2022-02-01Added 11 answers

Step 1

You can easily track where the extra solutions were introduced

$x+1+\sqrt{4x+1}=0\to {0+1+\sqrt{1}=0}$

$x+1=-\sqrt{4x+1}\to {0+1=-\sqrt{0+1}}$

$x}^{2}+2x+1=4x+1\to {0+0+1=0+1$

$x}^{2}-2x=0\to {0-0=0$

$x(x-2)=0\to {0(0-2)=0}$

Repeat with$x=2$ .

If you turn$a=b$ to $a}^{2}={b}^{2$ , you also allow $a=-b$ .

You can easily track where the extra solutions were introduced

Repeat with

If you turn