Department of Mathematics --- College of Science --- University of Utah

Mathematics 1010 online

Quadratic Equations

A quadratic equation can be written in the standard form

$\displaystyle ax^2+bx+c = 0 \qquad\qquad(*) $

where $ x $ is the unknown variable, the coefficients $ a $, $ b $ and $ c $ are real numbers, and

$\displaystyle a\neq 0. $

(If $ a=0 $ we have a linear equation . and a much simpler problem.) In a class like this the coefficients are usually given explicitly, but in actual applications they are often variables, or even algebraic expressions.

To solve the quadratic equation means to find values of $ x $ that make the equation true. To illustrate the principles and issues let's look at a special case first. Consider the equation

$\displaystyle (x-1)^2 = 25.\qquad\qquad(**) $

You might say that's not a quadratic equation since it's not in the form $ (*) $. However, it can be converted into an equivalent equation that is in that form, by suitable operations on both sides of the equation:

\begin{displaymath} \begin{array}{rcccl} (x-1)^2 &=& 25 &\vert& \hbox{expand} \\... ...24 &=& 0 &\vert& a=1,\quad b=-2, \quad c =-24.\\ \end{array} \end{displaymath}

The last equation in this sequence is in standard form, with $ a $, $ b $, and $ c $ having the given values. However, the equation $ (**) $ can be solved much more easily than $ (*) $:

\begin{displaymath} \begin{array}{rcccl} (x-1)^2 &=& 25 &\vert& \sqrt{\phantom{\... ...d\hbox{or}\quad{-4} &\vert& \hbox{the answer} \\ \end{array} \end{displaymath}

Thus there are two solutions of the equation, $ x=6 $ and $ x=-4 $. We can (and should) verify this by substituting these values in the original equation. If $ x=6 $ we obtain $ 5^2=25 $ and if $ x=-4 $ we obtain $ (-5)^2 = 25 $.

Note the symbol $ \pm $ in the second and third of the above sequence of equations. The square root of $ 25 $ is positive by convention and equals $ +5 $. However, our task at that stage is not to compute a square root as such, but to answer the question for what values of $ x $ does $ (x-1)^2 $ equal $ 25 $? There are two such values, $ x-1=-5 $ and $ x-1=-5 $, and we must consider both possibilities.

Let's now consider the more general equation

$\displaystyle (x-r)^2 = s \qquad\qquad(***) $

where $ r $ and $ s $ are considered known and, as before, $ x $ needs to be determined.

It can be solved just like the special case considered earlier:

\begin{displaymath} \begin{array}{rcccl} (x-r)^2 &=& s &\vert& \sqrt{\phantom{\l... ...x &=& r \pm \sqrt{s} &\vert& \hbox{the answer}\\ \end{array} \end{displaymath}

The key to solving quadratic equations is to convert them to the form $ (***) $. This process is called completing the square. It is based on the first and second binomial formulas .

Let's see how this works with our equation in standard form:

$\displaystyle x^2-2x-24 =0. $

If the constant term was $ 1 $ instead of $ -24 $ we would have a perfect square . To make it so we just add $ 25 $ on both sides and obtain

$\displaystyle x^2-2x+1 =25 $

which can be rewritten as

$\displaystyle (x-1)^2 = 25 $

Note that since

$\displaystyle (x-r)^2 = x^2-2rx+r^2 $

we simply look at the factor of $ x $, halve it, square it, and add the appropriate constant that makes the constant equal to that desired value. One simple minded way to obtain that constant is to subtract whatever constant is there, and then add the desired value. For this to work the leading coefficient (multiplying $ x^2 $) must equal $ 1 $. If it doesn't we divide first by the leading coefficient on both sides. (However, do not memorize this procedure as a recipe. Rather think about it, and practice it, until it makes sense and is so compelling that it becomes a natural part of your repertoire.)

Two Real Solutions

An example illustrates the process:

\begin{displaymath} \begin{array}{rclcr} 2x^2-10x+12 &=& 0 &\vert& \div 2 \\ x... ...uad\hbox{or}\quad 3 &\vert& \hbox{the answer.}\\ \end{array} \end{displaymath}

We easily check that $ x=2 $ and $ x=3 $ do in fact satisfy the original equation.

The above example illustrates one of three possible outcomes of this procedure, the case where there are two real solutions.

One Real Solution

In the following example there is only one solution:

\begin{displaymath} \begin{array}{rclcr} x^2+2x+1 &=&0 &\vert& \hbox{perfect squ... ...t& -1 \\ x &=& -1 &\vert& \hbox{the answer} \\ \end{array} \end{displaymath}

The reason there is only one solution is the fact that there is one and only one number whose square is $ 0 $.

Two Conjugate Complex Solutions

If the above procedure leads to taking the square root of a negative number we obtain a conjugate complex pair of solutions as illustrated in the following example:

\begin{displaymath} \begin{array}{rclcr} x^2 +2x + 2 &=& 0 &\vert& -2 \\ x^2 +... ...x{or}\quad x = -1-i &\vert& \hbox{the answer} \\ \end{array} \end{displaymath}

The Quadratic Formula

Of course there is nothing to stop us from applying this procedure to the general equation $ (*) $. This gives rise to the quadratic formula . Personally, I prefer not to burden my mind with having to memorize reliably yet another formula, and so I complete the square almost every time I solve a quadratic equation.