## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 8 - Section 8.2 - The Quadratic Formula - 8.2 Exercises: 36

#### Answer

$x=\left\{ \dfrac{-3-i\sqrt{7}}{4},\dfrac{-3+i\sqrt{7}}{4} \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $z(2z+3)=-2 ,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} z(2z)+z(3)=-2 \\\\ 2z^2+3z=-2 .\end{array} Using the properties of equality, the given equation is equivalent to \begin{array}{l}\require{cancel} 2z^2+3z+2=0 .\end{array} The quadratic equation above has $a= 2 , b= 3 , c= 2 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-3\pm\sqrt{3^2-4(2)(2)}}{2(2)} \\\\ x=\dfrac{-3\pm\sqrt{9-16}}{4} \\\\ x=\dfrac{-3\pm\sqrt{-7}}{4} .\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} x=\dfrac{-3\pm\sqrt{-1}\cdot\sqrt{7}}{4} .\end{array} Since $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=\dfrac{-3\pm i\sqrt{7}}{4} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{-3-i\sqrt{7}}{4} \\\\\text{OR}\\\\ x=\dfrac{-3+i\sqrt{7}}{4} .\end{array} Hence, $x=\left\{ \dfrac{-3-i\sqrt{7}}{4},\dfrac{-3+i\sqrt{7}}{4} \right\} .$

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