Completing the Square
Fall 2009
Completing the square takes an expression having a quadratic
term (x2), a linear term (x), and possibly a constant term, and
produces an equal expression involving squaring an expression (e.g. (x–3)2)
but no linear term.
For example,
completing the square for x2 – 4x + 15 gives (x–2)2 + 11.
It is simple to
check that these are equal because (x–2)2 + 11 = (x2 – 4x
+ 4) + 11 = x2 – 4x + 15.
Completing the
square tells us how to start with x2 – 4x + 15 and get (x–2)2 +
11. More generally, completing the square
takes an expression of the form ax2 + bx + c and finds an equal
expression of the form a(x – h)2 + k.
Completing the
square is often taught as one method to solve quadratic equations. (Other techniques are factoring and using
the quadratic formula.) Completing the
square also can be used to find the vertex of a parabola. In the above example, the parabola y
= x2 – 4x + 15 must have vertex (2, 11) because we
can rewrite the equation as y = (x–2)2 + 11 or
as y – 11 = (x–2)2. Other uses of completing the square will arise from time to time
in calculus.
Note that
completing the square is done on an expression, not an equation. So never try to add a number to "both
sides" since you are not dealing with an equality.
Here is a summary
of how to complete the square on various expressions.
Part 1
- Completing the square when x2
has coefficient 1, if there is no constant term.
- First look at the linear term and determine what
constant is needed to complete a perfect square. Do this by dividing the linear
coefficient by 2 (which is the same as multiplying by ½) and
then squaring.
- E.g. for x2 + 10x,
the desired constant term is found by dividing 10 by 2 and then
squaring: [½(10)]2 =
52 = 25
- Taking your original expression, add and
subtract the desired constant.
Since we are both adding and subtracting it, we get an expression
equal to the original.
- E.g. x2 +
10x = x2 + 10x + 25 – 25
- Factor the first three terms as a perfect square
using A2 ± 2AB + B2 = (A
± B)2.
- E.g. x2 +
10x + 25 – 25 = (x2 + 10x + 25) –25
= (x + 5)2 – 25
Here are some more examples; these will also be used below for more calculations:
x2 – 4x = x2 – 4x +
4 – 4 = (x2 – 4x + 4) – 4 = (x–2)2 – 4
x2 + 12x = x2 + 12x +
36 – 36 = (x2 + 12x + 36) – 36 = (x+6)2 – 36
x2 – 2x = x2 – 2x +
1 – 1 = (x2 – 2x + 1) – 1 = (x–1)2 – 1
x2 + x = x2 + x +
¼ – ¼ = (x2 + x + ¼) – ¼
= (x+½)2 – ¼
- Completing the square when x2
has coefficient 1, if there is a constant term.
- EITHER do the above procedure on the first two
terms, as side work, and then add the constant OR use the above procedure
directly.
- E.g. for x2
+ 10x + 7, we computed
above that x2 + 10x = (x + 5)2
– 25, and this allows us to
compute x2 + 10x + 7 = (x + 5)2 – 25 + 7 = (x + 5)2
– 18
- Or: x2 +
10x + 7 = x2
+ 10x + 25 – 25 + 7 = (x2+10x+25)–25+7
= (x + 5)2 – 18
More examples:
x2 – 6x + 15 = x2 – 6x +
9 – 9 + 15 = (x2 – 6x + 9) + (–9 + 15) = (x–3)2
+ 6
x2 +2x – 5 = x2 +2x +
1 – 1 – 5 = (x2 +2x + 1) + (–1 – 5) = (x+1)2
– 6
x2 + x + 2 = x2 + 3x +
¼ – ¼ + 2 = (x2 + x + ¼) + (–¼ + 2) = (x+½)2 + 7/4
Part 2
- Completing the square when the
coefficient of x2 is not 1:
Once again, you may choose to use side work. Here is how that works. - Factor out the coefficient of x2
from the quadratic and linear terms only, i.e. not from the
constant term. (The new
coefficients can be found by dividing.)
- E.g. 3x2 –
12x + 7 = 3(x2 – 4x) + 7
- As side work, complete the square on the terms
in parentheses.
- E.g. x2 – 4x = x2 – 4x + 4 – 4 = (x2 – 4x + 4) – 4 = (x–2)2 – 4
- Use this completed
square in the factored expression from the first step.
- E.g. 3(x2 –
4x) + 7 = 3[(x–2)2 – 4] + 7
- Simplify by
distributing. Distribute
carefully!
- E.g.
3[(x–2)2 – 4] + 7 = 3(x–2)2 – 12 + 7 = 3(x–2)2 –
5
Other examples which use our earlier calculations above as side work:
10x2 – 40x
– 2 = 10(x2 – 4x) – 2 = 10[(x–2)2 – 4] – 2
= 10(x–2)2 – 40 – 2 = 10(x–2)2 – 42
2x2 + 24x
= 2(x2 + 12x) = 2[(x+6)2 – 36] = 2(x+6)2
– 72
–x2 + 2x
+ 4 = –1(x2 – 2x) + 4 = –1[(x–1)2 – 1] + 4
= –1(x–1)2 + 1 + 4 = –(x–1)2 + 5
–3x2 – 3x
+ 2 = –3(x2+x)+2 = –3[(x+½)2 – ¼] + 2 = –3(x+½)2 + ¾ + 2 = –3(x+½)2 + 11/4
Examples done
without side work:
7x2 + 28x
+ 3 = 7(x2 + 4x) + 3 = 7(x2 + 4x +4
– 4) + 3 = 7[(x + 2)2 – 4] + 3 = 7(x
+ 2)2 – 28 + 3 = 7(x + 2)2 – 25
–2x2 + 20x
– 50 = –2(x2 – 10x) – 50 = –2(x2 – 10x +25
– 25) – 50 = –2[(x – 5)2 – 25] – 50 = –2(x
– 5)2 + 50 – 50 = –2(x – 5)2
Complete the square
on the following expressions. Check
your answers by multiplying out your result to see if you get the original
expression.
Exercises:
On exercises #1–10:
(a) Complete the
square. (Optional: check by multiplying out.)
(b) Sketch each function;
if possible, graph intercepts.
Hint: to graph, recall y=(x–h)2+k
is the standard parabola, shifted right h units and up k.
- x2 + 16x
- x2 – 14x
- x2 + 4x + 5
- x2 – 2x + 15
- x2 – 8x + 11
- x2 + 10x + 16
- x2 – 7x
- x2 + 5x
- x2 – 5x + 8
- x2 + 7x + 10
On Exercises
#11–29: Complete the square (you can
check by multiplying).
- 10t2 – 60t
- 8m2 + 16m
- 3y2 + 18y – 20
- 5w4 – 20w2 + 9
- 2c4 – 6c2 + 14
- –3t2 + 12t + 17
- –5p2 + 10p – 70
- 3k2 – 9k + 200
- –6j2 + 18j
- –5a2 – 25a
- 3q2 + 2q – 5
- 5n2 – 4n – 1
- ax2 + bx + c
There is a separate
Web page where you may check answers to
odd numbered problems.
Revised Friday, May 1, 2009. E-mail corrections, suggestions to mmaltenfort@ccc.edu