## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter P - Section P.7 - Equations - Exercise Set - Page 106: 86

#### 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, $2x^{2}+11x-6=0\quad\rightarrow \left\{\begin{array}{l} a=2\\ b=11\\ c=-6 \end{array}\right.$ $D=b^{2}-4ac=11^{2}-4(2)(-6)=121+48=169$ $D$ is positive $\Rightarrow$ two distinct real solutions.

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