Answer
no solutions
Work Step by Step
$\sqrt {x^2-7}+3=0$
$\sqrt {x^2-7}+3-3=0-3$
$\sqrt {x^2-7}=-3$
$(\sqrt {x^2-7})^2=(-3)^2$
$x^2-7=9$
$x^2=16$
$\sqrt {x^2} = \sqrt {16}$
$x=±4$
$\sqrt {x^2-7}+3=0$
$\sqrt {(-4)^2-7}+3=0$
$\sqrt {16-7}+3=0$
$\sqrt {9}+3=0$
$3+3=0$
$6=0$ (false)
$\sqrt {x^2-7}+3=0$
$\sqrt {(4)^2-7}+3=0$
$\sqrt {16-7}+3=0$
$\sqrt {9}+3=0$
$3+3=0$
$6=0$ (false)
