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: 18

Answer

$p=\left\{ \dfrac{-1-\sqrt{7}}{6},\dfrac{-1+\sqrt{7}}{6} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ p^2+\dfrac{p}{3}=\dfrac{1}{6} ,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the properties of equality, in the form $ax^2+bx+c=0,$ the expression above is equivalent to \begin{array}{l}\require{cancel} p^2+\dfrac{p}{3}-\dfrac{1}{6}=0 \\\\ 6\left(p^2+\dfrac{p}{3}-\dfrac{1}{6}\right)=6(0) \\\\ 6p^2+2p-1=0 .\end{array} The quadratic equation above has $a= 6 , b= 2 , c= -1 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} p=\dfrac{-2\pm\sqrt{2^2-4(6)(-1)}}{2(6)} \\\\ p=\dfrac{-2\pm\sqrt{4+24}}{12} \\\\ p=\dfrac{-2\pm\sqrt{28}}{12} .\end{array} Simplifying the radical by writing the radicand as an expression that contains a factor that is a perfect power of the index and then extracting the root of that factor result to \begin{array}{l}\require{cancel} p=\dfrac{-2\pm\sqrt{4\cdot7}}{12} \\\\ p=\dfrac{-2\pm\sqrt{(2)^2\cdot7}}{12} \\\\ p=\dfrac{-2\pm2\sqrt{7}}{12} .\end{array} Cancelling the common factors from all the terms results to \begin{array}{l}\require{cancel} p=\dfrac{\cancel2(-1)\pm\cancel2(1)\sqrt{7}}{\cancel2(6)} \\\\ p=\dfrac{-1\pm\sqrt{7}}{6} .\end{array} The solutions are \begin{array}{l}\require{cancel} p=\dfrac{-1-\sqrt{7}}{6} \\\\\text{OR}\\\\ p=\dfrac{-1+\sqrt{7}}{6} .\end{array} Hence, $ p=\left\{ \dfrac{-1-\sqrt{7}}{6},\dfrac{-1+\sqrt{7}}{6} \right\} .$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.