1. ## Another radical equation question

Solve the equation. Check for extraneous solutions.

$
\sqrt{2x + 3} + 2 = \sqrt{6x + 7}
$

I get to

$
0 = 4x
$

Does that mean no solution?

2. Originally Posted by happydino1
Solve the equation. Check for extraneous solutions.

$
\sqrt{2x + 3} + 2 = \sqrt{6x + 7}
$

I get to

$
0 = 4x
$

Does that mean no solution?
$\left( \sqrt{2x+3}+2 \right)^2 = 6x+7$
$2x+3 + 4 + 4\sqrt{2x+3} = 6x+7$
$4\sqrt{2x+3} = 4x$
$\sqrt{2x+3}=x$
$2x+3 = x^2$
Can you solve that?