Intermediate Algebra (12th Edition)

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

Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - 8.1 Exercises: 67

Answer

$x=\left\{ \dfrac{5-\sqrt{15}}{5},\dfrac{5+\sqrt{15}}{5} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 5x^2-10x+2=0 ,$ use first the properties of equality to express the equation in the form $x^2\pm bx=c.$ Once in this form, complete the square by adding $\left( \dfrac{b}{2} \right)^2$ to both sides of the equal sign. Then express the left side as a square of a binomial while simplify the right side. Then take the square root of both sides (Square Root Property) and use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Using the properties of equality, in the form $x^2+bx=c,$ the given equation is equivalent to \begin{array}{l}\require{cancel} \dfrac{5x^2-10x+2}{5}=\dfrac{0}{5} \\\\ x^2-2x+\dfrac{2}{5}=0 \\\\ x^2-2x=-\dfrac{2}{5} .\end{array} In the equation above, $b= -2 .$ The expression $\left( \dfrac{b}{2} \right)^2,$ evaluates to \begin{array}{l}\require{cancel} \left( \dfrac{-2}{2} \right)^2 \\\\= \left( -1 \right)^2 \\\\= 1.\end{array} Adding the value of $\left( \dfrac{b}{2} \right)^2,$ to both sides of the equation above results to \begin{array}{l}\require{cancel} x^2-2x+1=-\dfrac{2}{5}+1 \\\\ x^2-2x+1=-\dfrac{2}{5}+\dfrac{5}{5} \\\\ x^2-2x+1=\dfrac{3}{5} .\end{array} With the left side now a perfect square trinomial, the equation above is equivalent to \begin{array}{l}\require{cancel} \left( x-1 \right)^2=\dfrac{3}{5} .\end{array} Taking the square root of both sides (Square Root Property), simplifying the radical and then isolating the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} x-1=\pm\sqrt{\dfrac{3}{5}} \\\\ x-1=\pm\sqrt{\dfrac{3}{5}\cdot\dfrac{5}{5}} \\\\ x-1=\pm\sqrt{\dfrac{1}{25}\cdot15} \\\\ x-1=\pm\dfrac{1}{5}\sqrt{15} \\\\ x-1=\pm\dfrac{\sqrt{15}}{5} \\\\ x=1\pm\dfrac{\sqrt{15}}{5} \\\\ x=\dfrac{5}{5}\pm\dfrac{\sqrt{15}}{5} \\\\ x=\dfrac{5\pm\sqrt{15}}{5} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{5-\sqrt{15}}{5} \\\\\text{OR}\\\\ x=\dfrac{5+\sqrt{15}}{5} .\end{array} Hence, $ x=\left\{ \dfrac{5-\sqrt{15}}{5},\dfrac{5+\sqrt{15}}{5} \right\} .$
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