**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.