Question

Answer the question

Answer 1

We can do anything to an equation as long as we do it to both sides

given: $\sqrt{3+\frac{(4+\sqrt{x+3})^2}{6}}=3$
square both sides
$3+\frac{(4+\sqrt{x+3})^2}{6}=9$
minus 3 from both sides
$\frac{(4+\sqrt{x+3})^2}{6}=6$
multiply both sides by 6
$(4+\sqrt{x+3})^2=36 square root both sides, remember to take the positive and negative roots [tex]4+\sqrt{x+3}=+/-6$
minus 4
$\sqrt{x+3}=-4+/-6$
square both sides
do for -4+6 and -4-6 (2 and -10)
$x+3=4$ and $x+3=100$
minus 3
$x=1$ and $x=97$
test for extraneous roots
x=97 doesn't work


the answer is x=1