# Chapter 11 - Quadratic Functions and Equations - 11.1 Quadratic Equations - 11.1 Exercise Set: 27

$y=\dfrac{3\pm\sqrt{17}}{4}$

#### Work Step by Step

Since $x^2=a$ implies $x=\pm\sqrt{a}$ (the Square Root Principle), then the solutions to the given equation, $\left( y+\dfrac{3}{4} \right)^2=\dfrac{17}{16} ,$ are \begin{array}{l}\require{cancel} y+\dfrac{3}{4}=\pm\sqrt{\dfrac{17}{16}} \\\\ y+\dfrac{3}{4}=\pm\dfrac{\sqrt{17}}{\sqrt{16}} \\\\ y+\dfrac{3}{4}=\pm\dfrac{\sqrt{17}}{\sqrt{(4)^2}} \\\\ y+\dfrac{3}{4}=\pm\dfrac{\sqrt{17}}{4} \\\\ y=\dfrac{-3}{4}\pm\dfrac{\sqrt{17}}{4} \\\\ y=\dfrac{-3\pm\sqrt{17}}{4} .\end{array}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.