Chapter 6 - Radical Functions and Rational Exponents - 6-5 Solving Square Root and Other Radical Equations - Practice and Problem-Solving Exercises - Page 395: 35

$3$

Square both sides: $(\sqrt{3x})^2=(\sqrt{x+6})^2$ $3x=x+6$ Subtract $x$ from both sides: $3x-x=x+6-x$ $2x=6$ Divide both sides by $2$: $\dfrac{2x}{2}=\dfrac62$ $x=3$ Check for extraneous solutions: $\sqrt{3(3)}=\sqrt{(3)+6}$ $\sqrt{9}=\sqrt{9}$ $3=3$