# Chapter 10 - Exponents and Radicals - 10.6 Solving Radical Equations - 10.6 Exercise Set - Page 668: 54

$$x=10$$

#### Work Step by Step

We solve the equation using the rules of exponents and radicals. Note, if we wanted to check our solution, we could plug the value we get back into the original equation to see if it yields a true statement. Solving the equation, we find: $$\left(2\sqrt{3x+6}\right)^2=\left(5+\sqrt{4x+9}\right)^2\\ 12x+24=4x+10\sqrt{4x+9}+34\\ \left(8x-10\right)^2=\left(10\sqrt{4x+9}\right)^2\\ 64x^2-160x+100=400x+900 \\ x=10$$ Note, $x=1.25$ is not a valid solution.

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