# Chapter 9 - Roots and Radicals - 9.5 - Solving Radical Equations - Problem Set 9.5 - Page 424: 19

a=6

#### Work Step by Step

Recall, in order to cancel out a square root, we square both sides of the equation. Thus, we square both sides of the equation to obtain that 3a-2=2a+4. We now simplify: $a-2 =4 \\\\ a=6$ We now must check our solution: $\sqrt{3(6)-2}=\sqrt{2(6)+4} \\\\ \sqrt{16}=\sqrt{16} \\\\ 4=4$ This is true, so our solution is valid.

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