Hi everyone! Welcome back to integralcalc.com.
Today we are going to do our first example of the method of completing the square
which is one method that you can use to solve for the roots of polynomial equations
that can't easily be factored.
Let's go ahead and take a look at our first example.
It’s going to be:
f(x) = x^2 - 4x + 2
And the first thing that we are going to do in completing the square
is set this equation equal to zero.
So we just take the function exactly as it is and set it as equal to zero.
We are just going to walk through the steps of completing the square.
It's a fairly easy concept.
It's just a matter of memorizing the steps
to get you to the final answer, so, we set it equal to zero
then we take half of this "-4" term here.
Now, completing the square is a method that we can use
to solve for the roots of polynomials in the form
ax^2 + bx + c.
You can see that our polynomial here, we have:
ax^2 + bx + c
The "a" in this case is the coefficient on the "x^2" term which on this is "1",
"+bx", you have "-4x", so "b" is "-4"
and then "+c", we have "+2", so "c" is equal to "2".
So, our polynomial is in the form:
ax^2 + bx + c
where a, b, and c represent constants on those variable terms.
So, we take the
we take half of b,
the coefficient “b” on that x term which in our case is "-4",
so we just divide it by 2 and the answer that we get is "-2"
and that's the second thing that you'll always do in completing the square.
Our next step is to take our answer that we just got, this "-2",
and square it,
so we take (-2)^2 and we get 4.
Now 4 is the
number that completes the square of our first two terms,
so we're going to take that and we're going to add it into our original function
and subtract it at the same time so that our function,
our original function, remains essentially unchanged
but we can actually do something more with it.
So, you can see here we took the "4" from our previous answer
and we added it in to our original function
right after the first two terms,
x^2 - 4x,
so, we added the 4, but then, we also subtracted here
so that we could take those two out and our original function would be
unchanged.
So, we haven't change the function but we do this more workable because now
we can take the square of this first
of this first group of terms and parenthesis here.
Since we've gone to the steps of completing the square,
this is now the square of the term "x-2".
So, we go ahead and factor that and we get (x-2)^2
and then we just combine this "-4" and "+2"
into "-2" here, so we simplify.
And then,
once we've done that we go ahead and move 2 to the opposite side
and adding it to both sides,
we're moving towards solving for the roots of the original function.
So, our next step is to take the square root of both sides of the function.
We are trying to get x on its own
so we take the square roots so that we can remove this
this squared exponent here,
so we take the square root of both sides
and then that leaves us with "x-2" on one side
and the square root of 2 on the other side but what we need to remember is that
we can get
positive or negative square root of 2 on the right hand side.
You could have the negative square root of 2
and if you would square it you'd get a "+2"
or the positive square root of 2, which if you squared it would get 2
and so, since we're looking for the square root of 2 here,
we have to always remember to add that "+/–" there.
Then finally to solve for x,
we add this 2 here to both sides
so we get x on its own on the left side
and our
and our roots,
we have 2 +/- the square root of 2,
which, since we added that "+/-" sign
we'll break into two roots for us,
so when we do that,
we get two roots for this polynomial equation,
the first, being 2 plus the square root of 2 and the second, being 2 minus the square root of 2.
So, those are the two roots of our function.
That’s our final answer.
I hope that helped and I will see you guys next time.
Bye!!!