## Algebra 2 (1st Edition)

Published by McDougal Littell

# Chapter 4 Quadratic Functions and Factoring - 4.5 Solve Quadratic Equations by Finding Square Roots - 4.5 Exercises - Skill Practice - Page 270: 30

#### Answer

The solutions are $2\sqrt{35}$ and $-2\sqrt{35}.$

#### Work Step by Step

$\displaystyle \frac{t^{2}}{20}+8=15\qquad$ ...add $-8$ to each side. $\displaystyle \frac{t^{2}}{20}=7\qquad$ ...multiply each side with $20$. $t^{2}=140\qquad$ ...take square roots of each side. $\sqrt{t^{2}}=\sqrt{140}\qquad$ ...simplify. ...When solving an equation of the form $x^{2}=s$ where $s>0$, we find both the positive and negative solutions. $t=\pm\sqrt{140}\qquad$ ...rewrite $140$ as a product of two factors so that one factor is a perfect square. ($140=4\cdot 35$) $t=\pm\sqrt{4}\cdot 35\qquad$ ...use the Product Property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ $t=\pm\sqrt{4}\cdot\sqrt{35}\qquad$ ...evaluate $\sqrt{4}$ ($\sqrt{4}=2$) $t=\pm 2\sqrt{35}$

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