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 - Monitoring Progress - Page 562: 13

Answer

The solution is $p=7$.

Work Step by Step

The given equation is $\Rightarrow p+1=\sqrt{7p+15}$ Square each side of the equation. $\Rightarrow (p+1)^2=(\sqrt{7p+15})^2$ Simplify. $\Rightarrow p^2+2p+1=7p+15$ Add $-7p-15$ to each side. $\Rightarrow p^2+2p+1-7p-15=7p+15-7p-15$ Simplify. $\Rightarrow p^2-5p-14=0$ Factor. $\Rightarrow (p-7)(p+2)=0$ Use zero product property. $\Rightarrow p-7=0$ or $p+2=0$ Solve for $p$. $\Rightarrow p=7$ or $p=-2$ Check $p=7$. $\Rightarrow p+1=\sqrt{7p+15}$ $\Rightarrow 7+1=\sqrt{7(7)+15}$ $\Rightarrow 8=\sqrt{49+15}$ $\Rightarrow 8=\sqrt{64}$ $\Rightarrow 8=8$ True. Check $p=-2$. $\Rightarrow p+1=\sqrt{7p+15}$ $\Rightarrow -2+1=\sqrt{7(-2)+15}$ $\Rightarrow -1=\sqrt{-14+15}$ $\Rightarrow -1=\sqrt{1}$ $\Rightarrow -1=1$ False Hence, the solution is $p=7$.
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