## Precalculus (6th Edition) Blitzer

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). --- Add $-2x+1$ to both sides, so we have $ax^{2}+bx+c=0.$ $3x^{2}-2x+1=0=0\quad\rightarrow \left\{\begin{array}{l} a=3\\ b=-2\\ c=1 \end{array}\right.$ $D=b^{2}-4ac=(-2)^{2}-4(3)(1)=4-12=-8$ D is negative $\Rightarrow$ two solutions, complex conjugates (not real).