>> Welcome to the Cypress College Math Review on Completing the Square.
Our first objective deals with the square root property.
Before we get to what the square root property is we want
to review first what a quadratic equation in standard form looks like.
Suppose a, b, and c are real numbers such that a does not equal to zero.
Then any equation that could be written
in the form ax squared plus bx plus c equals zero is considered
to be a quadratic equation in standard form.
The quadratic part is because you have a term that has its variable with an exponent of two.
What makes it standard form is that it's equaling zero.
So with that in mind let's now look at what is the square root property.
Suppose k is a real number.
If x squared equals k, then x equals the square root of k or x equals negative square root
of k. This may also be written as x equals plus or minus the square root of k.
So we will use the plus or minus symbol to represent the positive
and the negative version of our square root.
While we still need to simplify the radical we now have some notation that's a little bit
more compact.
Now let us solve some quadratic equations actually using the square root property.
In our first example we have x squared plus 3 equals negative 15.
Before you can actually use the square root property you need
to isolate the square term first.
So here x squared is that square term and in order to get it alone on the left hand side
of the equation we need to subtract three form both sides.
This leaves us with x squared equaling negative 18.
Once we have x squared alone we are now ready to use the square root property,
which says x sill be equal to plus or minus the square root of negative 18.
Once we have done all this we can now focus on simplifying the radical.
One of the first things we can do when it comes to simplifying the square root
of negative 18 is remember that i equals the square root of negative 1.
So we can actually have that negative inside of our radical leave as i.
So we can now rewrite our statement as x equals plus or minus i times the square root of 18.
Now we can focus all of our attention on just simplifying the square root of 18.
Our goal is to try to find what perfect squares make up the number 18.
So let us simplify the radical by noticing x equals plus
or minus i times the square root of 2 times 9.
The reason why we may want to show our statement this way is so that we can see
that we have the perfect square nine inside of our square root and we know
since the square root of 9 is 3 we can simplify by saying x equals plus
or minus 3 i times the square root of 2.
Since we can't simplify any further we're now ready to write our solution set
which can be written as an open brace, and from here we just list our solutions that we found.
So we write negative 3 i times the square root of 2 and positive 3 i times the square root
of 2, and then we close it with a brace.
Let's try another example.
In this one we're going to notice that our square term is no longer just a single letter,
it actually contains an expression so let's read what we have to look at.
We have here the quantity 4 m plus 5 being squared, then we subtract 8 equaling 20.
So our goal is still the same as it was before; we can't use the square root property
until we isolate the squared term.
Here 4 m plus 5 is the quantity that is being squared so that is what we want to get alone.
First step we need to add 8 to both sides.
This leaves us with 4 m plus 5 quantity squared equals 28.
Here we can finally use the square root property which has the inside
of what was being squared staying the same,
4 m plus 5 equals plus or minus the square root of 28.
Before we focus our attention on solving for m on the left hand side
of the equation let's simplify the radical first, this will just allow it
so that we don't forget to do it later on and we're going to have
to do it anyway so let's just do it right now.
Let's keep 4 m plus 5 on the left equaling plus or minus the square root of 4 times 7.
Why we're writing 28 in this way is because we notice 4 is a perfect square
because the square root of 4 is 2, leaving the 7 left over inside of our square root.
So now we can rewrite our statement as 4 m plus 5 equals plus
or minus 2 times the square root of 7.
Since we can't simplify the radical any further we can now focus the rest
of our attention on solving for the variable.
Here the first thing we need to do will be to subtract 5 from both sides
so we get 4 m equaling negative 5 plus or minus 2 times the square root of 7.
After that we want to divide both sides of our equation by the number 4,
which leaves us with m equals negative 5 plus
or minus 2 times the square root of 7 all being divided by 4.
Once we have this we are now ready to write our final answer as a solution set.
So our solution set will take the form of open brace negative 5 minus 2 times the square root
of 7 all divided by 4, and we have our second solution
of negative 5 plus 2 times the square root of 7 all divided by 4; then we close it with a brace.
Pause the video and try these problems.
In objective two we're going to review perfect square trinomials.
A perfect square trinomial is a trinomial which is a three term polynomial
that when you factor it results in a binomial being squared.
So if you have a trinomial of the form a squared plus 2 ab plus b square it will factor
as a binomial a plus b quantity squared.
If instead of addition for your middle term you actually have subtraction the factorization
changes to a minus b quantity squared.
The way that you can remember this is that the sign in the middle
of the trinomial always match the sign in the middle of your binomial when it's factored.
Let us try some examples.
Factor the perfect square trinomial.
In part a we have x squared plus 18 x plus 81.
Well we first want to verify does this even have the structure of a perfect square trinomial?
How we can go about doing this is setting up some parenthesis that actually shows the formula
for a perfect square trinomial and if it checks
out we can then factor it using those pieces that we found.
So let's begin by writing some parenthesis,
leave some space inside, and put a square on the outside.
Then we'll write a plus 2 times and we'll write two more sets of parenthesis,
leaving the insides blank for now, plus a set of parenthesis squared.
Now we're going to go in and fill all these blanks.
The first set of parenthesis we are going to ask ourselves what would I need to square in order
to get the first term of x squared in the original trinomial.
That would have to be a x because x when you square it gives you x squared.
Now before we check the middle I always encourage people to go
to the back and let's check the last term.
Let's see there that what would you have to square in order to get 81.
Well 9. So now that you see that the first term and the last term are matching
up let's take what was in those two sets of parenthesis and put them into the middle set
of parenthesis and we're going to check is it true that 2 times x times 9 does it equal 18 x,
and if the answer is yes like it is
in this example then what we have is actually a perfect square trinomial,
which means we can now factor this by taking our x, our 9, and since the sign in the middle
of the original trinomial was a plus sign we also will have a plus sign.
So now we can write our factorization as x plus 9 quantity squared.
For our next example we're not just going to have the leading coefficient being 1,
what happens if the leading coefficient is another number?
So in example b we have 9 c squared minus 66 c plus 121.
We're going to set up everything the same way as we did before but now it's just going
to be a matter of what are the pieces that we need to create that above trinomial.
So let's set up our actual parenthesis, so we equal a set of parenthesis squared.
Since we have subtraction of that 66 c we're going
to write minus 2 times we'll write our two sets of parenthesis,
plus a set of parenthesis squared.
Now we're going to check that first term.
In order to get 9 c squared wouldn't you have to use a 3 c?
To get that 121 for the last term wouldn't you have to square 11?
So now we can take 3 c, now we can take our 11, fill in the middle two sets of parenthesis
and check is it true that negative 2 times 3 c times 11 does it
in fact equal negative 66 c. Well negative 2 times 3 c would give you negative 6 c
and negative 6 c when you multiply by 11 does in fact give you negative 66 c.
So again we have a perfect square trinomial so we can take the 3 c and the 11,
we can write them now taking the subtraction with us to get 3 c minus 11 quantity squared.
We are now ready for completing a perfect square trinomial.
We can make x squared plus bx a perfect square trinomial by adding
to it 1/2 times b and then we square that result.
The b in this formula is coming from the number in front of x;
this is known as the linear coefficient.
This is only possible though when the leading coefficient, the number that's in front
of the x squared, is positive 1, so we will have to be aware of that when we start playing
with different examples in the future.
Let's start now with an example of this.
Find what number must be added to the expression in order to make it a perfect square trinomial.
Then factor the trinomial.
We have x squared minus 8 x. The b value we have is negative 8, so we will take 1/2 multiply it
by negative 8 and we will square that result.
We could jump straight to the simplified answer here but it's actually going
to benefit us by showing the middle step.
This will allow us to use a shortcut when it comes to that factoring step.
So let's take half of negative 8, which is negative 4,
we'll square it; so negative 4 squared equals 16.
Sixteen is the number that is going to complete the perfect square trinomial for our expression,
so we write plus 16, and now we're ready to factor.
But when it comes to factoring this is where we can take advantage
of that shortcut that I've mentioned earlier.
Here we can write the letter of the problem, which is x, and the number you need
to factor this will be the number we had before we squared it.
So since we had negative 4 we're going to write minus 4.
We will put this in parenthesis and then put a power of 2 on the outside.
So we have x minus 4 quantity squared.
If we really wanted to check this multiplying
out x minus 4 quantity squared will get us the x squared minus 8 x plus 16.
Pause the video and try these problems.
Objective three, completing the square when a equals 1.
In this example we're going to solve the quadratic equation by completing the square.
What we have is x squared minus 4 x minus 10 equals zero.
Since the leading coefficient is already 1 all we need
to do first is isolate the variable terms.
This means get x squared minus 4 x alone on one side of the equal sign.
Well in order to do this we need to add 10 to both sides.
So by doing so we end up getting x squared minus 4 x equaling 10.
Once we have the variable terms alone on one side we're ready
to do that completing the square step.
We want to try to figure out what number we need to add to our expression on the left in order
to make it a perfect square trinomial.
This is where we use 1/2 times b and we square that result.
So to complete the perfect square trinomial in this problem we take 1/2 and we multiply it
by the b value here, which is negative 4, and we will square our result.
In order to still use the shortcut we will show the simplifying step,
1/2 times negative 4 is just negative 2, which we will square to get positive 4.
Four is the number that we want to now add to both sides of our equation.
This leads us to x squared minus 4 x plus 4 equaling 10 plus 4.
From here we're ready to factor the perfect square trinomial that we created on the left
and all we need to do to the right is simplify the numbers.
To factor the perfect square trinomial we will write x,
and since it was negative 2 before we squared it that's what we'll write next, minus 2,
put parenthesis around this so that we can square the quantity, and what we will be equal
to is 10 plus 4, which is just 14.
Now that everything has been factored and simplified we are now ready
to use the square root property to solve for the variable.
To do this we write x minus 2 equals plus or minus the square root of 14.
Since 14 does not contain any perfect square factors we leave our radical
as the square root of 14.
All we need to do now is add 2 to both sides of our equation
so that we get x equals 2 plus or minus the square root of 14.
And our final step is to use solution sets to represent our answers.
So we write an open brace, 2 minus the square root of 14, we'll write our comma,
and then we write 2 plus the square root of 14, and close it with a brace.
Pause the video and try these problems.
Objective four, completing the square when a does not equal 1.
In this example we're going to solve the quadratic equation by completing the square
but you're going to notice that our leading coefficient is no longer a positive 1.
Here we have 2 x squared plus 3 x minus 52 equaling zero.
We're going to start of the same way, let's isolate the variable terms first.
To do this we'll just add 52 to both sides of our equation.
This leads us to 2 x squared plus 3 x equaling 52.
From here we want to turn 2 x squared into just 1 x squared; in order to do this we're going
to divide every term by the leading coefficient a in the case when a is not equaling 1.
So in our problem we're going to divide every term on both sides of the equation by 2.
So we'll divide 2 x squared by 2, we'll divide 3 x by 2, and we'll divide 52 by 2.
When we reduce each of the fractions as best we can we get x squared plus 3/2 x equaling 26.
Now we want to complete the perfect square trinomial and solve
for the variable as we did before.
This is where we're going to use 1/2 times b and then square the result.
So in our problem we're going to take 1/2, we multiply it by the b value 3/2,
and then we'll square that result.
Simplifying gets us 3/4 squared, and then once we square we get 9/16.
Nine sixteenths is now the number we're going to add to both sides of our equation,
so we write x squared plus 3/2 x plus 9/16 equals 26 plus 9/16.
Once we've done this we can now factor the perfect square trinomial that is on the left.
This is absolutely when we want to take advantage of our shortcut.
Here we will write x, and since we had positive 3/4 we write plus 3/4 quantity squared.
On the right we are going to have to do some fraction addition where we will need
to find a common denominator and then rewrite the 26 plus 9/16 as a single fraction.
After we have done this work we end up getting 425/16.
We are now ready to use the square root property.
To do it we're going to get x plus 3/4 equals plus
or minus the square root of 425 divided by 16.
Before we move the 3/4 over, as always, let us simplify the radical first
so that we don't forget to do it.
We have a property of radicals that says whenever you have a quotient inside
of a radical you can split it apart so that you're taking the root
of both the numerator and denominator separately.
The reason why we do this is because we want
to make sure we don't have any radicals left over in a denominator.
In our problem we're going to have x plus 3/4 equals plus or minus the square root
of 425 divided by the square root of 16.
We can now simplify the numerator to get 5 times the square root of 17
and we can simplify the denominator to get 4.
This leaves us with x plus 3/4 equals plus
or minus 5 times the square root of 17 divided by 4.
We will now solve for x by subtracting 3/4 from both sides of our equation.
This leads to x equals negative 3/4 plus
or minus 5 times the square root of 17 divided by 4.
With x being solved we can now write our final solution set.
So we have negative 3/4 minus 5 times the square root of 17 divided by 4
and we have negative 3/4 plus 5 times the square root of 17 divided by 4.
Pause the video and try these problems.