# POLQ 2 | Lesson 3 | Solutions (Solving Radical Equations)

Solutions:

1.  $$\sqrt{2x-1} -5 = -2$$  →   $$\sqrt{2x-1} = 5-2$$→    $$\sqrt{2x-1} = 3$$ → $$(\sqrt{2x-1})^2 = 3^2$$→$$2x-1 =9$$→ $$2x =10$$ →$$x =5$$.     Five is  a solution to the equation.
2.  $$\sqrt{2x-1} +5 = 2$$→   $$\sqrt {2x-1} = -5+2$$→  $$\sqrt{2x-1} = -3$$→  $$(\sqrt {2x-1})^2 = (-3)^2$$→$$2x-1 =9$$→$$2x =10$$→$$x =5$$    Five is not a solution to this equation because $$8\neq2$$.  So there is not solution to this problem.

Always check to make sure the solution satisfies the equation.