Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.4 Equations - R.4 Exercises - Page R-16: 13


$-\displaystyle \frac{1}{4}$ and $\displaystyle \frac{2}{3}$

Work Step by Step

Set the $RHS$ to 0 by subtracting 2 from both sides $12x^{2}-5x=2\qquad.../-$2 $12x^{2}-5x-2=0$ To factor the trinomial, we search for factors of $12\times(-2)=-24, $ whose sum is $-5:\qquad $ We find $-8$ and $+3$. $12x^{2}-5x-2=12x^{2}-8x+3x-2$ ... and factor in pairs: $=4x(3x-2)+1(3x-2)$ $=(4x+1)(3x-2)$ Our equation becomes $(4x+1)(3x-2)=0$ By the zero product principle, $4x+1=0$ or $3x-2=0$ $4x=-1$ or $3x=2$ $x=-\displaystyle \frac{1}{4}$ or $x=\displaystyle \frac{2}{3}$ The solutions are $-\displaystyle \frac{1}{4}$ and $\displaystyle \frac{2}{3}$.
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.