THE QUADRATIC FORMULA

Quadratic functions are functions that can be written in the form [beautiful math coming... please be patient] $\,y = ax^2 + bx + c\,$ for $\,a\ne 0\,$.
Every quadratic function graphs as a parabola with directrix parallel to the $\,x$-axis.

The graphs of quadratic functions can have three different $\,x$-intercept situations, as shown below:
no [beautiful math coming... please be patient] $\,x$-intercepts;
exactly one $\,x$-intercept;
two different $\,x$-intercepts.

no [beautiful math coming... please be patient] $\,x\,$-intercepts exactly one $\,x\,$-intercept two different $\,x\,$-intercepts

Points on the [beautiful math coming... please be patient] $\,x$-axis have their $\,y$-value equal to zero.
Thus, to find the $\,x$-intercepts for any curve, you set $\,y\,$ equal to zero and solve for $\,x\,$.

In particular, to find the $\,x$-intercepts of a quadratic function [beautiful math coming... please be patient] $\,y = ax^2 + bx + c\,$   ($\,a\ne 0\,$),
it is necessary to solve the equation [beautiful math coming... please be patient] $\,ax^2 + bx + c = 0\,$.

The formula that gives the solutions to this equation is called the quadratic formula, and is derived next:

DERIVATION OF THE QUADRATIC FORMULA
(solving the equation [beautiful math coming... please be patient] $\,ax^2 + bx + c = 0\,$, where $\,a\ne 0\,$)

Using the technique of completing the square,
the equation is transformed to the form [beautiful math coming... please be patient] $\,z^2 = k\,$, as follows:

[beautiful math coming... please be patient] $\displaystyle ax^2 + bx + c = 0$ original equation
[beautiful math coming... please be patient] $\displaystyle ax^2 + bx = -c$ subtract $\,c\,$ from both sides
[beautiful math coming... please be patient] $\displaystyle x^2 + \frac{b}{a}x = -\frac{c}{a}$ divide both sides by $\,a\ne 0\,$;
the technique of completing the square only works
when the coefficient of the squared term is $\,1\,$
[beautiful math coming... please be patient] $\displaystyle x^2 + \frac{b}{a}x + {(\frac{b}{2a})}^2 = -\frac{c}{a} + \frac{b^2}{4a^2}$ notice that $\,\frac{b/a}{2} = \frac{b}{a}\div 2 = \frac{b}{a}\cdot \frac{1}{2} = \frac{b}{2a}\,$;
add [beautiful math coming... please be patient] $\,{(\frac{b}{2a})}^2 = \frac{b^2}{4a^2}\,$ to both sides;
this is the correct number to add
to turn the left side into a perfect square
[beautiful math coming... please be patient] $\displaystyle{\left(x + \frac{b}{2a}\right)}^2 = -\frac{c}{a}\cdot \frac{4a}{4a} + \frac{b^2}{4a^2}$ rewrite left side as a perfect square;
get a common denominator on the right
[beautiful math coming... please be patient] $\displaystyle{\left(x + \frac{b}{2a}\right)}^2 = \frac{b^2 - 4ac}{4a^2}$ add fractions, simplify
[beautiful math coming... please be patient] $\displaystyle x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}$ the equation $\,z^2 = k\,$ has solutions [beautiful math coming... please be patient] $\,z=\pm\sqrt{k}\ $;
recall that $\,z=\pm\sqrt{k}\,$ is just a convenient shorthand for
$\,(z = \sqrt{k}\ \ \text{or}\ \ z = -\sqrt{k})$
[beautiful math coming... please be patient] $\displaystyle x + \frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}$ use the rule [beautiful math coming... please be patient] $\displaystyle \,\sqrt{\frac{A}{B}} = \frac{\sqrt{A}}{\sqrt{B}}\,$
[beautiful math coming... please be patient] $\displaystyle x + \frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{\sqrt{4}\sqrt{a^2}}$ use the rule [beautiful math coming... please be patient] $\displaystyle \,\sqrt{AB} = \sqrt{A}\cdot\sqrt{B}\,$
[beautiful math coming... please be patient] $\displaystyle x + \frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{2|a|}$ use the rule $\,\sqrt{A^2} = |A|\,$

Notice that if [beautiful math coming... please be patient] $\,a\gt 0\,$, then $\,|a|=a\,$, so the right side of the previous equation becomes [beautiful math coming... please be patient] $\,\pm\frac{\sqrt{b^2-4ac}}{2a}\,$.

Also, if $\,a\lt 0\,$, then $\,|a|=-a\,$, so the right side becomes [beautiful math coming... please be patient] $\,\pm\frac{\sqrt{b^2-4ac}}{2(-a)} = \mp\frac{\sqrt{b^2-4ac}}{2a}\,$.

In both cases, the same two values for the right side result.
Thus, we continue with a simplified right side:

[beautiful math coming... please be patient] $\displaystyle x + \frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{2a}$ the plus/minus allows elimination of the absolute value
[beautiful math coming... please be patient] $\displaystyle x=-\frac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$ add $\displaystyle\,-\frac{b}{2a}\,$ to both sides
[beautiful math coming... please be patient] $\displaystyle x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$ add fractions

In summary, we have:

THE QUADRATIC FORMULA
Let $\,a\ne 0\,$.
The solutions to the equation $\,ax^2 + bx + c = 0\,$ are given by: [beautiful math coming... please be patient] $$ x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} $$

The expression [beautiful math coming... please be patient] $\,b^2 - 4ac\,$ that appears under the square root in the quadratic formula
is critical in determining the nature of the solutions to the quadratic equation $\,ax^2 + bx + c = 0\,$, as follows:

Thus, the expression [beautiful math coming... please be patient] $\,b^2 - 4ac\,$ helps us to discriminate between the various types of solutions to a quadratic equation,
and the various $\,x\,$-intercept situations for a quadratic function.

Thus, we have the following definition:

DEFINITION discriminant
Let [beautiful math coming... please be patient] $\,a\ne 0\,$.

The expression [beautiful math coming... please be patient] $\,b^2 - 4ac\,$ is called the discriminant of the quadratic equation [beautiful math coming... please be patient] $\,ax^2 + bx + c = 0\,$.

Similarly, the expression $\,b^2 - 4ac\,$ is called the discriminant of the quadratic function $\,ax^2 + bx + c\,$.
Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Complex Numbers


On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
(MAX is 14; there are 14 different problem types.)