Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.7 - Equations - Exercise Set - Page 106: 90

Answer

Two distinct real solutions.

Work Step by Step

For $ax^{2}+bx+c=0$, we can find solutions using the Quadratic formula: $\displaystyle \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. The radicand in the formula is called the discriminant. $D=b^{2}-4ac$ D is positive $\Rightarrow$ there are two distinct real solutions. D is zero $\Rightarrow$ there is one (double) real solution. D is negative $\Rightarrow$ two solutions, complex conjugates (not real). --- Here, $ 3x^{2}+4x-2=0\quad\rightarrow \left\{\begin{array}{l} a=3\\ b=4\\ c=-2 \end{array}\right.$ $D=b^{2}-4ac=4^{2}-4(3)(-2)=16+24=40$ D is positive $\Rightarrow$ two distinct real solutions.
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.