﻿ The Quadratic Formula

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

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} x &= \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 &= \frac{-14\pm\sqrt{256}}{6}\cr\cr &= \frac{-14\pm 16}{6} \end{align} Substitute into the quadratic formula and simplify: $$x = \frac{-\color{blue}{b}\pm\sqrt{\color{blue}{b}^2 - 4\color{red}{a}\color{green}{c}}}{2\color{red}{a}}$$
$\displaystyle x = \frac{-14 \,\color{orange}{\large{\mathbf{+}}} \,16}{6} = \frac{2}{6} = \frac 13$

or

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

Write the ‘$\,\color{orange}{\large{\mathbf{+}}}\,$’ and ‘$\,\color{purple}{\large{\mathbf{-}}}\,$’ solutions separately.
 $$\begin{gather} 14(\frac 13) + 3\bigl(\frac 13\bigr)^2 \overset{?}{=} 5\cr\cr \frac{14}{3} + \frac{1}{3} \overset{?}{=} 5\cr\cr \frac{15}{3} \overset{?}{=} 5\cr\cr 5 = 5 \end{gather}$$ $$\begin{gather} \cr 14(-5) + 3\bigl(-5)^2 \overset{?}{=} 5\cr\cr -70 + 75 \overset{?}{=} 5\cr\cr 5 = 5 \end{gather}$$
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.
Voila!

## 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 $\,\frac{b/a}{2} = \frac{b}{a}\div 2 = \frac{b}{a}\cdot \frac{1}{2} = \frac{b}{2a}\,$; 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: $$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

## 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:

• If $\,b^2 - 4ac\gt 0\,$, then $\sqrt{b^2-4ac}\,$ is a positive real number.
In this case,   $\displaystyle\,\frac{-b + \sqrt{b^2-4ac}}{2a}\,$   and   $\displaystyle\,\frac{-b - \sqrt{b^2-4ac}}{2a}\,$   are different real numbers.
Thus, there are two different real number solutions to the quadratic equation $\,ax^2 + bx + c = 0\,$,
and the graph of the quadratic function $\,ax^2+bx+c\,$ has two different $\,x\,$-intercepts.
• If $\,b^2 - 4ac= 0\,$, then $\sqrt{b^2-4ac} = 0\,$.
In this case,   $\displaystyle\,\frac{-b + \sqrt{b^2-4ac}}{2a} = \frac{-b+0}{2a}\,$   and   $\displaystyle\,\frac{-b - \sqrt{b^2-4ac}}{2a} = \frac{-b-0}{2a}\,$   are the same numbers.
Thus, there is exactly one real number solution to the quadratic equation $\,ax^2 + bx + c = 0\,$,
and the graph of the quadratic function $\,ax^2 + bx + c\,$ has only one $\,x\,$-intercept.
• If $\,b^2 - 4ac\lt 0\,$, then   $\sqrt{b^2-4ac}\,$   is not a real number.
In this case, there are no real number solutions to the quadratic equation $\,ax^2 + bx + c = 0\,$,
and the graph of the quadratic function $\,ax^2+bx+c\,$ has no $\,x\,$-intercepts.

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.
PROBLEM TYPES:
 1 2 3 4 5 6 7 8 9 10 11 12 13 14
AVAILABLE MASTERED IN PROGRESS
 (MAX is 14; there are 14 different problem types.)