## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Section 1.5 - Equations - 1.5 Exercises: 56

#### Answer

$x=\dfrac{\sqrt{7}-1}{5}$ $x=-\dfrac{1+\sqrt{7}}{5}$

#### Work Step by Step

$(5x+1)^{2}+3=10$ We need to solve this by factoring, so let's take the $10$ to the left side of the equation to substract: $(5x+1)^{2}+3-10=0$ Simplify: $(5x+1)^{2}-7=0$ Remember the difference of two squares factoring formula: $a^{2}-b^{2}=(a-b)(a+b)$ In this equation $a=(5x+1)^{2}$ and $b=7$, so we can factor it and we get: $[(5x+1)-\sqrt{7}][(5x+1)+\sqrt{7}]=0$ Then, we set each factor equal to $0$ and solve them as individual equations. Let the $[(5x+1)-\sqrt{7}]=0$ be equation number 1 and $[(5x+1)+\sqrt{7}]=0$ be equation number 2: Solving for $x$ in equation number 1: $(5x+1)-\sqrt{7}=0$ $5x=\sqrt{7}-1$ $x=\dfrac{\sqrt{7}-1}{5}$ Solving for $x$ in equation number 2: $(5x+1)+\sqrt{7}=0$ $5x=-1-\sqrt{7}$ $x=-\dfrac{1+\sqrt{7}}{5}$

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