Why do I get incorrect solutions x=0 and x=2 for x+1+4x+1=0?

Question

basgrwthej · Accepted Answer

Step 1
Here is one correct way to solve this problem I learnt from comments.
I should write

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

⟺ {\begin{cases} (x + 1)^{2} = 4 x + 1 \\ (x + 1) \leq 0 \end{cases}

⟺ {\begin{cases} x (x - 2) = 0 \\ x \leq - 1 \end{cases}

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 \leq 1

. Therefore the equation has no solutions.

Karly Logan · Accepted Answer

Step 1
You can easily track where the extra solutions were introduced

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

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

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

x^{2} - 2 x = 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

.

Why do I get incorrect solutions x=0 and x=2 for