Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 8 - Section 8.2 - The Quadratic Formula - 8.2 Exercises: 13

Answer

$x=\left\{ \dfrac{-1-\sqrt{2}}{2},\dfrac{-1+\sqrt{2}}{2} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ 4x^2+4x-1=0 ,$ express in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ The quadratic equation above has $a= 4 , b= 4 , c= -1 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-4\pm\sqrt{4^2-4(4)(-1)}}{2(4)} \\\\ x=\dfrac{-4\pm\sqrt{16+16}}{8} \\\\ x=\dfrac{-4\pm\sqrt{32}}{8} .\end{array} Simplifying the radical by writing the radicand as an expression that contains a factor that is a perfect power of the index and then extracting the root of that factor result to \begin{array}{l}\require{cancel} x=\dfrac{-4\pm\sqrt{16\cdot2}}{8} \\\\ x=\dfrac{-4\pm\sqrt{(4)^2\cdot2}}{8} \\\\ x=\dfrac{-4\pm4\sqrt{2}}{8} .\end{array} Cancelling the common factors from all the terms results to \begin{array}{l}\require{cancel} x=\dfrac{\cancel4(-1)\pm\cancel4(1)\sqrt{2}}{\cancel4(2)} \\\\ x=\dfrac{-1\pm\sqrt{2}}{2} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{-1-\sqrt{2}}{2} \\\\\text{OR}\\\\ x=\dfrac{-1+\sqrt{2}}{2} .\end{array} Hence, $ x=\left\{ \dfrac{-1-\sqrt{2}}{2},\dfrac{-1+\sqrt{2}}{2} \right\} .$
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