Exercise 57 Page 588

Let's analyze the Quadratic Formula. The discriminant is the expression under the radical sign, $b^2-4ac.$ We know that the discriminant can be used to determine the number of real solutions of a quadratic equation in the following way.

Discriminant	Number of Real Solutions
Negative	None
Zero	One
Positive	Two

We will explain why the discriminant can be used to determine the number of real zeros.

Negative Discriminant

Let's think what would happen if the discriminant is negative, $\boxed{b^2-4ac}<0.$ Then we have to calculate a square root of a $\text{negative}$ $\text{number},$ which is impossible for real numbers. Therefore, the Quadratic Formula does not make sense in this context. This tells us that the quadratic equation does not have any real zeros.

Zero Discriminant

Now, we will analyze what we get when $\boxed{b^2-4ac}=0.$ Let's substitute $0$ for the discriminant in the Quadratic Formula. In this case, we find exactly one real solution.

Positive Discriminant

Finally, let's assume that $\boxed{b^2-4ac}>0.$ In this case, we calculate a square root of a positive number. Each positive real number has exactly one real root. This means that the solutions for the equation are $x=\frac{\text{-}b\pm \sqrt{b^2-4ac}}{2a}.$ Let's separate them into the positive and negative cases.

$x=\frac{\text{-}b\pm \sqrt{b^2-4ac}}{2a}$
$x_1=\frac{\text{-}b-\sqrt{b^2-4ac}}{2a}$	$x_2=\frac{\text{-}b+\sqrt{b^2-4ac}}{2a}$

Therefore, we have exactly two real solutions, $x_1$ and $x_2.$