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: 65

Answer

$x=\left\{ \dfrac{-5-\sqrt{41}}{4},\dfrac{-5+\sqrt{41}}{4} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 2k^2+5k-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{2k^2+5k-2}{2}=\dfrac{0}{2} \\\\ k^2+\dfrac{5}{2}k-1=0 \\\\ k^2+\dfrac{5}{2}k=1 .\end{array} In the equation above, $b= \dfrac{5}{2} .$ The expression $\left( \dfrac{b}{2} \right)^2,$ evaluates to \begin{array}{l}\require{cancel} \left( \dfrac{\dfrac{5}{2}}{2} \right)^2 \\\\= \left( \dfrac{5}{2}\div{2} \right)^2 \\\\= \left( \dfrac{5}{2}\cdot\dfrac{1}{2} \right)^2 \\\\= \left( \dfrac{5}{4} \right)^2 \\\\= \dfrac{25}{16} .\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} k^2+\dfrac{5}{2}k+\dfrac{25}{16}=1+\dfrac{25}{16} \\\\ k^2+\dfrac{5}{2}k+\dfrac{25}{16}=\dfrac{16}{16}+\dfrac{25}{16} \\\\ k^2+\dfrac{5}{2}k+\dfrac{25}{16}=\dfrac{41}{16} .\end{array} With the left side now a perfect square trinomial, the equation above is equivalent to \begin{array}{l}\require{cancel} \left( k+\dfrac{5}{4} \right)^2=\dfrac{41}{16} .\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} k+\dfrac{5}{4}=\pm\sqrt{\dfrac{41}{16}} \\\\ k+\dfrac{5}{4}=\pm\sqrt{\dfrac{1}{16}\cdot41} \\\\ k+\dfrac{5}{4}=\pm\dfrac{1}{4}\sqrt{41} \\\\ k+\dfrac{5}{4}=\pm\dfrac{\sqrt{41}}{4} \\\\ k=-\dfrac{5}{4}\pm\dfrac{\sqrt{41}}{4} \\\\ k=\dfrac{-5\pm\sqrt{41}}{4} .\end{array} The solutions are \begin{array}{l}\require{cancel} k=\dfrac{-5-\sqrt{41}}{4} \\\\\text{OR}\\\\ k=\dfrac{-5+\sqrt{41}}{4} .\end{array} Hence, $ x=\left\{ \dfrac{-5-\sqrt{41}}{4},\dfrac{-5+\sqrt{41}}{4} \right\} .$
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