# Chapter 11 - Quadratic Functions and Equations - 11.1 Quadratic Equations - 11.1 Exercise Set: 30

$x=\{-7,13\}$

#### Work Step by Step

Using $a^2+2ab+b^2=(a+b)^2$ or the factoring of perfect square trinomials, the given equation, $x^2-6x+9=100 ,$ is equivalent to \begin{array}{l}\require{cancel} (x-3)^2=100 .\end{array} Since $x^2=a$ implies $x=\pm\sqrt{a}$ (the Square Root Principle), the solutions to the equation, $(x-3)^2=100 ,$ are \begin{array}{l}\require{cancel} x-3=\pm\sqrt{100} \\\\ x-3=\pm\sqrt{(10)^2} \\\\ x-3=\pm10 \\\\ x=3\pm10 \\\\ x=\{-7,13\} .\end{array}

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