by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!
Thanks for your support!

A quadratic equation is an equation of the form $\,ax^2 + bx + c = 0\,$, where $\,a\ne 0\,$.

The quadratic formula solves quadratic equations.
Just read off $\,a\,$, $\,b\,$, and $\,c\,$, and substitute into the quadratic formula:

THE QUADRATIC FORMULA:       $\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

For example, solving the quadratic equation ‘$\,14x + 3x^2 = 5\,$’ is as simple as this:

$\,14x + 3x^2 = 5\,$ original equation
$\,\color{red}{3}x^2 + \color{blue}{14}x \color{green}{- 5} = 0\,$ put the equation in standard form
$\,\color{red}{a = 3}\,$

$\,\color{blue}{b = 14}\,$

$\,\color{green}{c = -5}\,$
After the equation is in standard form, read off:
  • $\,a\,$ (the coefficient of the $\,x^2\,$ term)
  • $\,b\,$ (the coefficient of the $\,x\,$ term)
  • $\,c\,$ (the constant term)
Be sure to include the appropriate plus or minus signs!
$\displaystyle \begin{align} \cssId{s21}{x}\ &\cssId{s22}{= \frac{-\color{blue}{14}\pm\sqrt{\color{blue}{14}^2 - 4\cdot \color{red}{3}\cdot (\color{green}{-5})}}{2\cdot \color{red}{3}}}\cr\cr &\cssId{s23}{= \frac{-14\pm\sqrt{256}}{6}}\cr\cr &\cssId{s24}{= \frac{-14\pm 16}{6}} \end{align} $ Substitute into the quadratic formula and simplify: $$ \cssId{s26}{x = \frac{-\color{blue}{b}\pm\sqrt{\color{blue}{b}^2 - 4\color{red}{a}\color{green}{c}}}{2\color{red}{a}}} $$
$\displaystyle \cssId{s27}{x} \cssId{s28}{= \frac{-14 \,\color{orange}{\large{\mathbf{+}}} \,16}{6}} \cssId{s29}{= \frac{2}{6}} \cssId{s30}{= \frac 13}$


$\displaystyle \cssId{s32}{x} \cssId{s33}{= \frac{-14\,\color{purple}{\large{\mathbf{-}}}\,16}{6}} \cssId{s34}{= \frac{-30}{6}} \cssId{s35}{= -5}$
The symbol ‘$\,\pm\,$’ means ‘plus or minus’.

Write the ‘$\,\color{orange}{\large{\mathbf{+}}}\,$’ and ‘$\,\color{purple}{\large{\mathbf{-}}}\,$’ solutions separately.
$$ \begin{gather} \cssId{s40}{14(\frac 13) + 3\bigl(\frac 13\bigr)^2 \overset{?}{=} 5}\cr\cr \cssId{s41}{\frac{14}{3} + \frac{1}{3} \overset{?}{=} 5}\cr\cr \cssId{s42}{\frac{15}{3} \overset{?}{=} 5}\cr\cr \cssId{s43}{5 = 5} \end{gather} $$ $$ \begin{gather} \cr \cssId{s44}{14(-5) + 3\bigl(-5)^2 \overset{?}{=} 5}\cr\cr \cssId{s45}{-70 + 75 \overset{?}{=} 5}\cr\cr \cssId{s46}{5 = 5} \end{gather} $$
If you've got time, check your solutions in the original equation.
Don't drop the question mark over the equal sign until you're sure they're equal.

Quadratic Functions and their Relationship to Quadratic Equations

Quadratic functions are functions that can be written in the form $\,y = ax^2 + bx + c\,$ 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 $\,x$-intercepts;
exactly one $\,x$-intercept;
two different $\,x$-intercepts.

no $\,x\,$-intercepts exactly one $\,x\,$-intercept two different $\,x\,$-intercepts

Points on the $\,x$-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 $\,y = ax^2 + bx + c\,$   ($\,a\ne 0\,$),
it is necessary to solve the equation $\,ax^2 + bx + c = 0\,$ (which is called a quadratic equation).

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 $\,ax^2 + bx + c = 0\,$ , where $\,a\ne 0\,$)

Using the technique of completing the square,
the equation is transformed to the form $\,z^2 = k\,$, as follows:

$\displaystyle ax^2 + bx + c = 0$ original equation
$\displaystyle ax^2 + bx = -c$ subtract $\,c\,$ from both sides
$\displaystyle x^2 + \frac{b}{a}x = -\frac{c}{a}$ divide both sides by $\,a\ne 0\,$;
the technique of completing the square only works
when the coefficient of the squared term is $\,1\,$
$\displaystyle x^2 + \frac{b}{a}x + {(\frac{b}{2a})}^2 = -\frac{c}{a} + \frac{b^2}{4a^2}$ notice that $\,\cssId{s79}{\frac{b/a}{2}} \cssId{s80}{= \frac{b}{a}\div 2} \cssId{s81}{= \frac{b}{a}\cdot \frac{1}{2}} \cssId{s82}{= \frac{b}{2a}}\,$;
add $\,{(\frac{b}{2a})}^2 = \frac{b^2}{4a^2}\,$ to both sides;
this is the correct number to add
to turn the left side into a perfect square
$\displaystyle{\left(x + \frac{b}{2a}\right)}^2 = -\frac{c}{a}\cdot \frac{4a}{4a} + \frac{b^2}{4a^2}$ rewrite left side as a perfect square;
get a common denominator on the right
$\displaystyle{\left(x + \frac{b}{2a}\right)}^2 = \frac{b^2 - 4ac}{4a^2}$ add fractions, simplify
$\displaystyle x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}$ the equation $\,z^2 = k\,$ has solutions $\,z=\pm\sqrt{k}\ $;
recall that $\,z=\pm\sqrt{k}\,$ is just a convenient shorthand for
$\,(z = \sqrt{k}\ \ \text{or}\ \ z = -\sqrt{k})$
$\displaystyle x + \frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}$ use the rule $\displaystyle \,\sqrt{\frac{A}{B}} = \frac{\sqrt{A}}{\sqrt{B}}\,$
$\displaystyle x + \frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{\sqrt{4}\sqrt{a^2}}$ use the rule $\displaystyle \,\sqrt{AB} = \sqrt{A}\cdot\sqrt{B}\,$
$\displaystyle x + \frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{2|a|}$ use the rule $\,\sqrt{A^2} = |A|\,$

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

Also, if $\,a\lt 0\,$, then $\,|a|=-a\,$, so the right side becomes $\,\pm\frac{\sqrt{b^2-4ac}}{2(-a)} = \mp\frac{\sqrt{b^2-4ac}}{2a}\,$.

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

$\displaystyle x + \frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{2a}$ the plus/minus allows elimination of the absolute value
$\displaystyle x=-\frac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$ add $\displaystyle\,-\frac{b}{2a}\,$ to both sides
$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ add fractions

In summary, we have:

Let $\,a\ne 0\,$.
The solutions to the equation $\,ax^2 + bx + c = 0\,$ are given by: $$ \cssId{s117}{x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}} $$

The Discriminant of a Quadratic Equation or Quadratic Function

The expression $\,b^2 - 4ac\,$ 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 $\,b^2 - 4ac\,$ 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 $\,a\ne 0\,$.

The expression $\,b^2 - 4ac\,$ is called the discriminant
of the quadratic equation $\,ax^2 + bx + c = 0\,$.

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.
1 2 3 4 5 6 7 8 9 10 11 12 13 14

(MAX is 14; there are 14 different problem types.)