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Thank you for visiting our site! You landed on this page because you entered a search term similar to this: online factoring of complex quadratic equations with variables only, here's the result:

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 evenalgebraic expressions.

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

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

You might say that's not a quadratic equation since it's not in theform $ (*) $.However, it can be converted into an equivalentequation 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 theoriginal 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 sequenceof equations. The square root of $ 25 $ is positive by convention andequals $ +5 $. However, our task at that stage is not to compute asquare root as such, but to answer the question for what values of$ 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 $ needsto 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 thesquare. 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 factorof $ x $, halve it, square it, and add the appropriate constant thatmakes the constant equal to that desired value. One simple minded wayto obtain that constant is to subtract whatever constant is there, andthen add the desired value. For this to work the leading coefficient(multiplying $ x^2 $) must equal $ 1 $. If it doesn't we divide firstby the leading coefficient on both sides. (However, do not memorizethis procedure as a recipe. Rather think about it, and practice it,until it makes sense and is so compelling that it becomes a naturalpart 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 theoriginal equation.

The above example illustrates one of three possible outcomes of thisprocedure, 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 oneand only one number whose square is $ 0 $.

Two Conjugate Complex Solutions

If the above procedure leads to taking the square root of a negativenumber we obtain a conjugate complex pair of solutions as illustratedin 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 procedureto thegeneral 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 almosteverytime I solve a quadratic equation.