## Precalculus (6th Edition)

Published by Pearson

# Chapter 1 - Equations and Inequalities - 1.4 Quadratic Equations - 1.4 Exercises - Page 121: 36

#### Answer

$\color{blue}{\left\{\dfrac{5-2i\sqrt2}{2}, \dfrac{5+2i\sqrt2}{2}\right\}}$.

#### Work Step by Step

RECALL: If $x^2=a$, then taking the square root of both sides gives $x = \pm \sqrt{a}$. Take the square root of both sides of the given equation to obtain: $\sqrt{(-2x+5)^2}=\pm \sqrt{-8} \\-2x+5 =\pm \sqrt{4(-1)(2)} \\-2x+5 =\pm \sqrt{2^2(-1)(2)} \\-2x+5 =\pm 2\sqrt{(-1)(2)}$ Since $\sqrt{-1}=i$, then the expression above is equivalent to: $-2x+5=\pm 2i\sqrt{2}$ Add $-5$ to both sides: $-2x =-5 \pm 2i\sqrt{2}$ Divide $-2$ to both sides: $x=\dfrac{-5\pm 2i\sqrt2}{-2} \\x=\dfrac{5\mp 2i\sqrt2}{2}$ Thus, the solution set is $\color{blue}{\left\{\dfrac{5-2i\sqrt2}{2}, \dfrac{5+2i\sqrt2}{2}\right\}}$.

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