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Solving Quadratic Equations by Completing the Square

Module by: Kenny M. Felder. E-mail the author

Summary: This module teaches the method of solving quadratic equations by completing the square.

Consider the equation:

x + 3 2 = 16 x + 3 2 = 16 size 12{ left (x+3 right ) rSup { size 8{2} } ="16"} {}
(1)

We can solve this by analogy to the way that we approached absolute value problems. something squared is 16. So what could the something be? It could be 4. It could also be 44 size 12{ - 4} {}. So the solution is:

  • x + 3 = 4 x + 3 = 4 size 12{x+3=4} {}
    • x = 1 x = 1 size 12{x=1} {}
  • x + 3 = 4 x + 3 = 4 size 12{x+3= - 4} {}
    • x = 7 x = 7 size 12{x= - 7} {}

These are the two solutions.

This simple problem leads to a completely general way of solving quadratic equations—because any quadratic equation can be put in a form like the above equation. The key is completing the square which, in turn, is based on our original two formulae:

x + a 2 = x 2 + 2 ax + a 2 x + a 2 = x 2 + 2 ax + a 2 size 12{ left (x+a right ) rSup { size 8{2} } =x rSup { size 8{2} } +2 ital "ax"+a rSup { size 8{2} } } {}
(2)
x a 2 = x 2 2 ax + a 2 x a 2 = x 2 2 ax + a 2 size 12{ left (x - a right ) rSup { size 8{2} } =x rSup { size 8{2} } - 2 ital "ax"+a rSup { size 8{2} } } {}
(3)

As an example, consider the equation x2+10x+21=0x2+10x+21=0 size 12{x rSup { size 8{2} } +"10"x+"21"=0} {}. In order to make it fit one of the patterns above, we must replace the 21 with the correct number: a number such that x2+10x+__x2+10x+__ size 12{x rSup { size 8{2} } +"10"x+"__"} {} is a perfect square. What number goes there? If you are familiar with the pattern, you know the answer right away. 10 is 5 doubled, so the number there must be 5 squared, or 25.

But how do we turn a 21 into a 25? We add 4, of course. And if we add 4 to one side of the equation, we have to add 4 to the other side of the equation. So the entire problem is worked out as follows:

Example 1

Table 1
Solving by Completing the Square (quick-and-dirty version)
Solve x2+10x+21=0x2+10x+21=0 size 12{x rSup { size 8{2} } +"10"x+"21"=0} {} The problem.
x 2 + 10 x + 25 = 4 x 2 + 10 x + 25 = 4 size 12{x rSup { size 8{2} } +"10"x+"25"=4} {} Add 4 to both sides, so that the left side becomes a perfect square.
x + 5 2 = 4 x + 5 2 = 4 size 12{ left (x+5 right ) rSup { size 8{2} } =4} {} Rewrite the perfect square.
x + 5 = 2 x = 3 x + 5 = 2 x = 3 size 12{ matrix { x+5=2 {} ## x= - 3 } } {} x + 5 = 2 x = 7 x + 5 = 2 x = 7 size 12{ matrix { x+5= - 2 {} ## x= - 7 } } {} If something-squared is 4, the something can be 2, or 2 2 size 12{ - 2} {} . Solve both possibilities to find the two answers.

Thus, we have our two solutions.

Completing the square is more time-consuming than factoring: so whenever a quadratic equation can be factored, factoring is the preferred method. (In this case, we would have factored the original equation as x+3x+7x+3x+7 size 12{ left (x+3 right ) left (x+7 right )} {} and gotten straight to the answer.) However, completing the square can be used on any quadratic equation. In the example below, completing the square is used to find two answers that would not have been found by factoring.

Example 2

Table 2
Solving by Completing the Square (showing all the steps more carefully)
Solve 9x254x+80=09x254x+80=0 size 12{9x rSup { size 8{2} } - "54"x+"80"=0} {} The problem.
9x 2 54 x = 80 9x 2 54 x = 80 size 12{9x rSup { size 8{2} } - "54"x= - "80"} {} Put all the x x size 12{x} {} terms on one side, and the number on the other
x 2 6x = 80 9 x 2 6x = 80 9 size 12{x rSup { size 8{2} } - 6x= - { {"80"} over {9} } } {} Divide both sides by the coefficient of x 2 x 2 size 12{x rSup { size 8{2} } } {} (*see below)
x 2 6x + 9 ̲ = 80 9 + 9 ̲ x 2 6x + 9 ̲ = 80 9 + 9 ̲ size 12{x rSup { size 8{2} } - 6x {underline {+9}} = - { {"80"} over {9} } + {underline {9}} } {} Add the same number (*see below) to both sides, so that the left side becomes a perfect square.
x 3 2 = 80 9 + 81 9 = 1 9 x 3 2 = 80 9 + 81 9 = 1 9 size 12{ left (x - 3 right ) rSup { size 8{2} } = - { {"80"} over {9} } + { {"81"} over {9} } = { {1} over {9} } } {} Rewrite the perfect square.
x 3 = 1 3 x = 3 1 3 x 3 = 1 3 x = 3 1 3 size 12{ matrix { x - 3= { {1} over {3} } {} ## x=3 { { size 8{1} } over { size 8{3} } } } } {} x 3 = 1 3 x = 2 2 3 x 3 = 1 3 x = 2 2 3 size 12{ matrix { x - 3= - { {1} over {3} } {} ## x=2 { { size 8{2} } over { size 8{3} } } } } {} If something-squared is 1 9 1 9 size 12{ { {1} over {9} } } {} , the something can be 1 3 1 3 size 12{ { {1} over {3} } } {} , or 1 3 1 3 size 12{ - { {1} over {3} } } {} . Solve both possibilities to find the two answers.

Two steps in particular should be pointed out here.

In the third step, we divide both sides by 9. When completing the square, you do not want to have any coefficient in front of the term; if there is a number there, you divide it out. Fractions, on the other hand (such as the 809809 size 12{ - { {"80"} over {9} } } {} in this case) do not present a problem. This is in marked contrast to factoring, where a coefficient in front of the x2x2 size 12{x rSup { size 8{2} } } {} can be left alone, but fractions make things nearly impossible.

The step after that is where we actually complete the square. x2+6x+__x2+6x+__ size 12{x rSup { size 8{2} } +6x+"__"} {} will be our perfect square. How do we find what number we want? Start with the coefficient of x2x2 size 12{x rSup { size 8{2} } } {} (in this case, 6). Take half of it, and square the result. Half of 6 is 3, squared is 9. So we want a 9 there to create x2+6x+9x2+6x+9 size 12{x rSup { size 8{2} } +6x+9} {} which can be simplified to x+32x+32 size 12{ left (x+3 right ) rSup { size 8{2} } } {}.

Figure 1
An image showing how to select the correct number and position it within the equation correctly when completing the square.

If the coefficient of xx size 12{x} {} is an odd number, the problem becomes a little uglier, but the principle is the same. For instance, faced with:

x 2 + 5x + __ x 2 + 5x + __ size 12{x rSup { size 8{2} } +5x+"__"} {}
(4)

You would begin by taking half of 5 (which is 5252 size 12{ { {5} over {2} } } {}) and then squaring it:

x 2 + 5x + 25 4 = x + 5 2 2 x 2 + 5x + 25 4 = x + 5 2 2 size 12{x rSup { size 8{2} } +5x+ { {"25"} over {4} } = left (x+ { {5} over {2} } right ) rSup { size 8{2} } } {}

Another “completing the square” example, in which you cannot get rid of the square root at all, is presented in the worksheet “The Generic Quadratic Equation.”

One final note on completing the square: there are three different possible outcomes.

  • If you end up with something like x32=16x32=16 size 12{ left (x - 3 right ) rSup { size 8{2} } ="16"} {} you will find two solutions, since x3x3 size 12{x - 3} {} can be either 4, or 44 size 12{ - 4} {}. You will always have two solutions if the right side of the equation, after completing the square, is positive.
  • If you end up with x32=0x32=0 size 12{ left (x - 3 right ) rSup { size 8{2} } =0} {} then there is only one solution: x must be 3 in this example. If the right side of the equation is 0 after completing the square, there is only one solution.
  • Finally, what about a negative number on the right, such as x32=16x32=16 size 12{ left (x - 3 right ) rSup { size 8{2} } = - "16"} {}? Nothing squared can give a negative answer, so there is no solution.

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