9.2 - The Completing-the-square method

9.2 - The completing-the-square method

The completing the square method takes a quadratic trinomial in which the variable, call it x, occurs twice (shown here in red) and rewrites it in such a way that x only occurs once:

The method is based on factoring perfect square quadratic trinomials. Look at the following three examples where three quadratic trinomials in standard form on the left have been converted to completed square form on the right:

(You should distribute the expressions on the right and verify that they do equal the expressions on the left.) The first example was chosen to be a perfect square quadratic trinomial. Notice that its completed square form is exactly the same as its factored form.

In the second example the trinomial’s additive constant, 15, is 6 bigger than the additive constant, 9, in the first example. Notice that in the completed square form this surplus of 6 is just added on at the end.

Finally, in the third example the additive constant, 0, is 9 smaller than the additive constant in the first example. In the completed square form this deficit of 9 is just subtracted off at the end.

Conclusion: The completed square form is just the perfect square in factored form with some remainder constant tagged on at the end.

Here is a recipe for converting a quadratic to completed square form. This recipe also explains what to do if the leading coefficient a of the quadratic is not equal to 1.

Convert a quadratic expression to completed square form by following these steps:

Start with the quadratic expression in standard form, i.e. with the terms in the order: quadratic, linear, constant:
a x² + b x + c
If the leading coefficient (i.e. the coefficient of the quadratic term) is not equal to 1 then factor it out. This may produce fractions in the linear and constant terms but the important thing is to get a leading coefficient of 1 inside the brackets:
From now on we ignore the factor outside the brackets and continue working on the quadratic trinomial inside the brackets. Take the coefficient of the linear term (namely b/a), divide it by 2, and then square it. This produces the quantity .
Add and immediately subtract back off the quantity inside the brackets, right after the linear term, like this:
Use brackets to group the first three terms inside the brackets:

The purpose of adding to this group was to make this group a perfect square. But to keep the value of the entire expression unchanged, the rest had to have subtracted from it.
Write the perfect square in factored form:
Write “the rest” as a single fraction. The result is the so-called completed square form:

Example: Convert the quadratic trinomial x² + 6 x − 5 to completed square form.

Solution: Follow these steps:

This quadratic trinomial has coefficients a = 1, b = 6, and c = −5. Since it is already in standard form and since the leading coefficient is 1 we can proceed straight to step 3. Take the coefficient of the linear term, divide it by 2, and then square it. This gives 9.
Add 9 and subtract 9 back off again just after the linear term, like this:
x² + 6 x + 9 − 9 − 5.
Use brackets to group the first 3 terms. The bracketed group had 9 added to it to make it a perfect square. The rest of the expression had 9 subtracted from it. Call it the rest:
(x² + 6 x + 9) − 9 − 5.
Factor the perfect square group and simplify the rest:
(x + 3)² − 14.

The quadratic trinomial is now in completed square form.

Example: Convert the quadratic trinomial 2 x² + 9 x − 20 to completed square form.

Solution: Just for variety we will complete the square on this one using floating point rather than exact arithmetic (i.e. decimal numbers rather than fractions). Follow these steps:

This quadratic trinomial has coefficients a = 2, b = 9, and c = −20. Since it is already in standard form and we can proceed straight to step 2. Factor out the coefficient of the quadratic term:
2 ( x² + 4.5 x − 10)
Now look inside the brackets. Take the coefficient of the linear term, divide it by 2, and then square it. This gives 5.0625.
Add 5.0625 and subtract 5.0625 back off again just after the linear term, like this:
2 (x² + 4.5 x + 5.0626 − 5.0625 − 10)
Use square brackets to group the first 3 terms. This square-bracketed group had 5.0625 added to it to make it a perfect square. The rest of the expression had 5.0625 subtracted from it. Call it the rest:
2 ( [x² + 4.5 x + 5.0626] − 5.0625 − 10)
Factor the perfect square group and simplify the rest:
2 ( [x + 2.25]² − 15.0625)

The quadratic trinomial is now in completed square form.

The following example is based on these facts about parabolas:

Parabolas: The general equation of a parabola is:

y = a x² + b x + c,

where a, b and c are constants. From the general equation the only thing that you can say about the parabola is that it opens upward if a is positive and opens downward if a is negative. If you complete the square on the right-hand-side of this equation you get the so-called standard equation of a parabola:

y = a(x − h)² + k,

From this equation you can say that the vertex of the parabola is located at (h, k).

Example: Given the parabola y = x² −2x + 10. Complete the square on the right side, find the vertex and then sketch the parabola.

Solution: Follow these steps to complete the square on the right side:

The quadratic trinomial on the right-hand-side has coefficients a = 1, b = −2, and c = 10. Since it is already in standard form and since the leading coefficient is 1 we can proceed straight to step 3. Take the coefficient of the linear term, divide it by 2, and then square it. This gives 1.
Add 1 and subtract 1 back off again just after the linear term, like this:
Use brackets to group the first 3 terms. The bracketed group had 1 added to it to make it a perfect square. The rest of the expression had 1 subtracted from it. Call it the rest:
Factor the perfect square group and simplify the rest:

This is the standard equation for this parabola. From it we can tell that the parabola opens upward and that its vertex is at (1, 9). Here is a sketch of it (the red curve).

The following example is based on these facts about circles:

Circles: The general equation of a circle is:

x² + y² + a x + b y + c = 0,

where a, b and c are constants. If you complete the square twice, once on the x terms and once on the y terms, you get the so-called standard equation of a circle:

(x − h)² + (y − k)² = r²

From this equation you can say that the center of the circle is at (h, k) and that the radius is r.

Example: Given the circle x² + y² − 2 x + 4 y + 1 = 0. Complete the square twice, once on the x terms and once on the y terms. Use the result to find the center and radius of the circle and then sketch it.

Solution: Write the two x terms, leave a space, write the two y terms, leave another space, and then write the rest of the equation, like this:

To complete the square on the x terms, look only at the linear and quadratic x terms and ignore all the other terms (shown greyed out). Take the coefficient of the linear term, divide it by 2, and then square it. This gives 1. Add 1 and subtract 1 back off again in the space just after the linear term (shown in red), like this:

To complete the square on the y terms, look only at the linear and quadratic y terms and ignore all the other terms (shown greyed out). Take the coefficient of the linear term, divide it by 2, and then square it. This gives 4. Add 4 and subtract 4 back off again in the space just after the linear term (shown in blue), like this:

Altogether, you now have this:

Terms 1 to 3 are a perfect square in x and terms 5 to 7 are a perfect square in y. Put brackets around both perfect squares:

Write both perfect squares in factored form:

Add the three remaining numbers (to get −4), move this to the right side, and write it as the square of something (namely of 2):

This is the standard equation for this circle. From it we can tell that the circle has center at (1, −2) and radius 2. Here is a sketch of it (the red circle).

Algebra Coach Exercises

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