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 - Page 519: 32


$t=\left\{ -2-i\sqrt{7},-2+i\sqrt{7} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ t^2+4t+11=0 ,$ use the Quadratic Formula. $\bf{\text{Solution Details:}}$ The quadratic equation above has $a= 1 , b= 4 , c= 11 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} t=\dfrac{-4\pm\sqrt{4^2-4(1)(11)}}{2(1)} \\\\ t=\dfrac{-4\pm\sqrt{16-44}}{2} \\\\ t=\dfrac{-4\pm\sqrt{-28}}{2} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the equation above is equivalent to \begin{array}{l}\require{cancel} t=\dfrac{-4\pm\sqrt{-1}\cdot\sqrt{28}}{2} .\end{array} Since $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} t=\dfrac{-4\pm i\sqrt{28}}{2} .\end{array} Simplifying the radicand by writing it as an expression that contains a factor that is a perfect square of the index and then extracting the root of that factor, the equation above is equivalent to \begin{array}{l}\require{cancel} t=\dfrac{-4\pm i\sqrt{4\cdot7}}{2} \\\\ t=\dfrac{-4\pm i\sqrt{(2)^2\cdot7}}{2} \\\\ t=\dfrac{-4\pm i(2)\sqrt{7}}{2} \\\\ t=\dfrac{-4\pm 2i\sqrt{7}}{2} .\end{array} Cancelling the common factor in each term results to \begin{array}{l}\require{cancel} t=\dfrac{\cancel2(-2)\pm \cancel2(1)i\sqrt{7}}{\cancel2(1)} \\\\ t=-2\pm i\sqrt{7} .\end{array} The solutions are \begin{array}{l}\require{cancel} t=-2- i\sqrt{7} \\\\\text{OR}\\\\ t=-2+i\sqrt{7} .\end{array} Hence, $ t=\left\{ -2-i\sqrt{7},-2+i\sqrt{7} \right\} .$
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