It’s Professor Dave, let’s complete the square.
We just learned how to solve polynomials by factoring, but not all polynomials can be
factored, so we definitely need a few other techniques.
One technique that will always work is called completing the square, so let’s see how
that works now.
First consider the following polynomial, X squared plus two X plus one.
This can be factored, but furthermore, it can be factored into two identical binomials.
The factors of one are one and one, or negative one and negative one, and the positive ones
do indeed add up to two, so we can factor this into X plus one times X plus one.
Anything times itself is by definition that thing squared, so we can also express this
as X plus one quantity squared.
If this were equal to zero, the solution is very simple, it’s just negative one, because
that’s the value for X that will make this equal to zero.
So the original polynomial must be a perfect square, since it can be generated by squaring
this binomial, just like four is a perfect square because it’s two squared.
This kind of thinking comes in handy for certain polynomials, because while they might not
factor, we can deliberately add some term to the expression that will manufacture a
polynomial that is a perfect square, thus allowing us to express it as the square of
a binomial, and making it easy to solve.
This will make more sense with an example.
Let’s say we have X squared plus two X minus six equals zero.
If we find the factors of negative six, which are combinations of one and six or two and
three, we find that there is no way to add any pair of factors together to get positive
two, so we can’t factor this polynomial.
But let’s do a little trick.
Let’s add six to both sides, bringing the six to the other side.
Earlier we saw that X squared plus two X plus one is a perfect square, so what if we now
add one so that we suddenly do have a perfect square polynomial, in essence completing the
square?
Well whatever we do, we have to do it to both sides, so we end up with X squared plus two
X plus one equals seven.
The reason this is useful, is that we can now express the left side as X plus one quantity
squared.
From here, it’s not very hard to solve.
We have an exponent of two, so we take the square root of both sides.
That gives us X plus one equals plus or minus root seven.
Remember that when taking the square root of a number, we get two answers, since both
of these square to give seven.
Now we subtract one to solve for X, and the right side can now be expressed as two values,
either root seven minus one, or negative root seven minus one.
While these are not as aesthetically pleasing as integers, they are indeed the two solutions
to this polynomial.
Let’s try another.
Say we have X squared plus six X minus two equals zero.
Once again, we would quickly see that this can’t be factored, so let’s bring two
to the other side.
Now, to complete the square, all we do is take this coefficient of X, in this case six,
and we must cut it in half and then square it.
Six divided by two is three, and then we square that to get nine.
We add nine to both sides to complete the square, leaving us with this expression.
The reason we do this, is because we need a C term with factors that multiply to give
C and add to give B. But since it’s a perfect square, the factors must be identical, so
that factor must be the square root of C, and at the same time add to itself, or double,
to give B. So when doing these problems, just cut the X coefficient in half to get the factor
that will go in the binomial, and then square it to get the C term you must add to both
sides, and that’s all you need to worry about.
Now we do have a perfect square, so we get X plus three quantity squared equals eleven.
Take the square root and subtract three, and we get X equals root eleven minus three or
negative root eleven minus three.
Let’s do just one more.
Three X squared plus fifteen X minus eight equals zero.
Once again we bring eight to the other side, but now we have something trickier on the
left.
The best thing to do when there is a coefficient on the X squared term is to divide everything
by that coefficient.
You may end up with fractions, but this is definitely the easiest way to go even if that’s
the case.
So let’s divide everything by three.
We will get X squared plus five X equals eight thirds.
Now we can complete the square.
With five, it may seem messy, because cutting in half gives five halves, which then squares
to give twenty-five fourths, but if that’s what we get, then that’s what we have to
work with.
We add that to the right side as well, find the least common denominator, which is twelve,
and convert these fractions to thirty-two twelfths and seventy-five twelfths so that
we can add them together to get a hundred and seven twelfths.
Now we factor the left side to get X plus five halves quantity squared.
Remember, when we cut that initial B term in half, before we square it, that’s the
number that will end up in the resulting binomial.
Then just like the other examples, we take the square root of both sides, and subtract
five halves, which will allow us to express the two solutions for X, being sure to simplify
radical expressions if applicable.
Don’t worry if the answers look unappealing, if that’s what we get, then that’s what
it is, and solutions won’t always be neat and tidy.
With this new technique under our belts, let’s check comprehension.