College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.4 - Quadratic Equations - 1.4 Exercises - Page 112: 69


$x=\left\{ -3,\dfrac{3-3i\sqrt{3}}{2},\dfrac{3+3i\sqrt{3}}{2} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ x^3+27=0 ,$ use the factoring of the sum or difference of $2$ cubes. Then solve the first factor using the properties of equality, while solve the second factor using the Quadratic Formula. $\bf{\text{Solution Details:}}$ The expressions $ x^3 $ and $ 27 $ are both perfect cubes (the cube root is exact). Hence, $ x^3+27=0 $ is a $\text{ sum }$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x)^3+(3)^3=0 \\\\ (x+3)(x^2-3x+9)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x+3=0 \\\text{ OR }\\ x^2-3x+9=0 .\end{array} Using the properties of equality, the solution to the first equation is \begin{array}{l}\require{cancel} x+3=0 \\\\ x=-3 .\end{array} In the second equation, $ x^2-3x+9=0 ,$ $a= 1 ,$ $b=-3 ,$ and $c= 9 .$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then \begin{array}{l}\require{cancel} x=\dfrac{-(-3)\pm\sqrt{(-3)^2-4(1)(9)}}{2(1)} \\\\ x=\dfrac{3\pm\sqrt{9-36}}{2} \\\\ x=\dfrac{3\pm\sqrt{-27}}{2} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy}$ and that $i=\sqrt{-1},$ the expression above is equivalent to\begin{array}{l}\require{cancel} x=\dfrac{3\pm\sqrt{-1}\cdot\sqrt{27}}{2} \\\\ x=\dfrac{3\pm i\cdot\sqrt{9\cdot3}}{2} \\\\ x=\dfrac{3\pm i\cdot\sqrt{(3)^2\cdot3}}{2} \\\\ x=\dfrac{3\pm 3i\sqrt{3}}{2} .\end{array} The solutions are $ x=\left\{ -3,\dfrac{3-3i\sqrt{3}}{2},\dfrac{3+3i\sqrt{3}}{2} \right\} .$
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