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


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.
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