Big Ideas Math - Algebra 1, A Common Core Curriculum

Published by Big Ideas Learning LLC
ISBN 10: 978-1-60840-838-2
ISBN 13: 978-1-60840-838-2

Chapter 10 - Radical Functions and Equations - 10.3 - Solving Radical Equations - Exercises - Page 564: 19

Answer

The solution is $p=15$.

Work Step by Step

The given equation is $\Rightarrow 7+3\sqrt{3p-9}=25$ Subtract $7$ from each side. $\Rightarrow 7+3\sqrt{3p-9}-7=25-7$ Simplify. $\Rightarrow 3\sqrt{3p-9}=18$ Divide each side by $3$. $\Rightarrow \sqrt{3p-9}=6$ Square each side of the equation. $\Rightarrow (\sqrt{3p-9})^2=(6)^2$ Simplify. $\Rightarrow 3p-9=36$ Add $9$ to each side. $\Rightarrow 3p-9+9=36+9$ Simplify. $\Rightarrow 3p=45$ Divide each side by $3$. $\Rightarrow p=15$ Check $p=15$. $\Rightarrow 7+3\sqrt{3p-9}=25$ $\Rightarrow 7+3\sqrt{3(15)-9}=25$ $\Rightarrow 7+3\sqrt{45-9}=25$ $\Rightarrow 7+3\sqrt{36}=25$ $\Rightarrow 7+3(6)=25$ $\Rightarrow 7+18=25$ $\Rightarrow 25=25$ True. Hence, the solution is $p=15$.
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