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# Solving Quadratic Equations by Completing the Square

Certainly an equation like is an easy equation to solve.  One need only take the square root of both sides to determine that the two solutions are .

Equations like  are slightly more complicated, but solved in a similar manner.  One would again take the square root of both sides to give the pair of equations  and .  By subtracting 3 from both sides in each one, we see that the  or .

So if we can get an equation into the form  , then the solutions must be of the form .  The method of completing the square does just that.  It is a way to modify a given quadratic equation so that it is of the form .

Example 1

Let’s assume we want to solve the equation:

Solution 1

The first step in completing the square is to group all the constant terms to one side of the equation.  In this case, we add 4 to both sides yielding the equation:

Then we ask ourselves, “How much must be added to both sides in order for the left hand side to be the square of a simple expression, like  .”  Perhaps you can see that the answer is to add 9 to the both sides, since .  If you didn’t, don’t worry.  The appropriate number to add is always the square of half of the coefficient of the x term.  In this example, the coefficient of x is 6.  Therefore, the number we must add to complete the square, is  which equals 9.

Careful, though!  Don’t forget the cardinal rule of algebra: “What you do to one side of an equation you must do to the other as well.”  So, if we add 9 to the left hand side, we must add 9 to the right hand side, too.  This gives us the equation:

Or more concisely,

Now we can solve as we did before, first getting x + 3 = 4 or x + 3 = -4, giving us the two solutions: and .  You may check to confirm that both of these are solutions to our original equation, .

If x = 1, then  (1)2 + 6(1) – 4 = 3.

If x = -7, then  (-7)2 + 6(-7) – 4 = 3.

Example 2

Things get a little more complicated if we introduce a coefficient of the term that is not 1.  Say for instance that we wanted to solve the equation .

Solution 2

We begin as in example 1 and group all the constant terms on one side of the equation.  We do this by subtracting 14 from both sides to give the equivalent equation:

The difference is in this next step.  We want the coefficient of to be 1, not 3.  So we simply divide both sides of the equation by 3, like this:

When simplified, we have the equation:

Now we once again halve the coefficient of x and square it to complete the square.  In this case we get which equals 4 (note the sign changes because we are squaring a negative).  Adding 4 to both sides of the equation gives us:

Or more concisely,

Then:                          (x – 2) = 5       or         (x – 2) = -5
x = 7                or         x = -3

Solving a quadratic equation by completing the square is sometimes a relatively easy task, although there are occasions when the coefficients can make it messy.  Fortunately, there is a no-fail method for solving quadratic equations called the quadratic formula.  Its application is the subject of another tutorial.