Help me solve for $x$ in the equation: $$\sqrt{2\sqrt{3x+3}}=\sqrt{2x+2}$$
Radical Equation Help Algebra
-4
$\begingroup$
algebra-precalculus
1 Answers
3
Assuming that the square root over the equation is just over the left side and you want to solve for $x$...
$$\sqrt{2\sqrt{3x+3}}=\sqrt{2x+2}$$ Squaring both sides $$2\sqrt{3x+3}=2x+2$$ Dividing by $2$ $$\sqrt{3x+3}=x+1$$ Squaring both sides again $$3x+3=x^2+2x+1$$ Simplifying $$x^2-x-2=0$$ Factoring $$(x-2)(x+1)=0$$ Solving for $x$ $$x=2\;\text{or}\;x=-1$$
Testing that the solutions work and no mischief occured with the squaring of the equations:
Substituting $2$ for $x$ gives you $$\sqrt{2\sqrt{9}}=\sqrt{6}$$ and simplifying leads to $$\sqrt6=\sqrt6\;\color{green}\checkmark$$
Substituting $-1$ for $x$ gives you $$\sqrt{2\sqrt{0}}=\sqrt{0}$$ and simplifying leads to $$0=0\;\color{green}\checkmark$$