>> We now have four methods for solving quadratic equations.
Let's talk about when to use which method.
So we have the method of taking square roots,
also called extracting square roots.
We have the method of factoring, solve by factoring.
We have the method of solve by completing the square
and we have solve by the quadratic formula.
So it's like having four different tools in your toolkit.
Let's talk about when to use which tool.
This first problem is ideal for solving by taking square roots
because the variable is inside the perfect square.
To solve this, we do the opposite
of your order of operations.
So we want to isolate the perfect square
by first adding 5 to both sides.
Next, we divide both sides by 2.
So we've isolated the perfect square.
Our next step is to take the square root of both sides.
Remember, there are two square roots,
so we have to put the plus or minus
in front of the square root.
Now, we want to simplify the square root of 12.
Since that is 2 times 2 times 3, we would end up 2 on the outside
and then the square root of 3.
Subtract 3 from both sides.
Those are not like terms on the right-hand side
so I cannot combine them into one.
You can write your answer like that or solution set,
we would write it like this.
Here's our answer.
For this problem we're going to solve by factoring.
So move all terms to one side.
So add 14x to both sides so it's going become positive.
So on the right side, we just have 0.
And then factor this trinomial.
There's lot of different ways to factor.
That's not what we're going over in this video.
Set each factor equal to 0.
Solve each equation that you get.
And then if you are asked to write your answer
in solution set, you would do so.
Solving by factoring is one of the easier methods.
This problem, we're going to complete the square.
So first, we move the constant term to the other side.
The constant is the number.
So in this case, the number is the negative 8.
So to move the constant to the other side, we're going to add 8
to both sides of the equation.
So that leaves us with a positive 8
on the right-hand side of the equation.
Our next step is to divide it by the leading coefficient.
The leading coefficient is the 4.
The 4 is being multiplied times x squared.
So we're going to divide each term of the equation by 4.
That leaves as with 1x squared.
To be able to complete the square, you need a 1x squared.
So now, I have it in the appropriate form to be able
to complete the square.
Now, we look at the number in front of x, we have to take half
of that number, square it and add it to both sides.
So half of 6 is 3, square is 9.
So we're going to add 9 to both sides of the equation.
The left-hand side becomes the perfect square, x plus 3,
remember that's a perfect square and that's equal to 11.
The next step is to take the square root of both sides.
Recall there are two square roots.
So we have x plus 3 equals plus or minus the square root of 11.
So x is equal to negative 3 plus or minus the square root of 11.
In solution set our answer would look like this.
For this problem, we're going to use the quadratic formula.
First, you move all terms to one side, so we're going
to add 12x to both sides.
Put them in their correct order.
Determine a, b, and c.
Write down the formula.
Substitute in your values.
And then simplify your result.
I usually use a calculator
for what's called the discriminant which is this part.
And then if you look at 60, 60 is equal to 4 times 15
which is 2 times 2 times 15.
So I can take 2 outside of the radical.
Yeah.
So I can factor that with 2.
And then I can cancel that 2
out of the numerator and the denominator.
And then if you are asked
to write your answer in solution set--
You have our answer.
So now you've had the chance to study each method.
We just went through
and reviewed one problem using each method.
And now we're going to go into a study
of which method is best for each problem.
When would you use one method?
When would you use another method?
What method is easiest for a particular problem?
So first of all, let's make sure you understand the terms.
In a quadratic equation,
the ax squared part is the quadratic term.
The bx part is the linear term, a line, liner.
The c part is the constant, the number, OK?
So when I say quadratic, that's the x squared part.
Linear, that's the x of the first part.
And constant, that's the number.
If your equation does not have a constant,
then solve by factoring.
So if it does not have a constant term,
move all terms to one side.
And since it doesn't have a constant,
every single term has an x in it.
We'll factor that x out.
So it's going to be a very easy equation to solve
like the equation I have here.
It will easily factor, set each factor equal to 0
and you've got it solved.
So if your equation does not have a constant term,
then definitely solving
by factoring will be the easiest method to solve that equation.
In general, any time an equation factors easily,
then that is definitely going to be the easiest method
to solve that equation.
If your equation does not have a linear term,
then solve by taking square roots.
Note here that we have a quadratic term, 4x squared,
and a constant term, negative 9, but there's no linear term.
So your goal here is to get the perfect square by itself.
The perfect square is just x square.
You want to get x squared by itself.
So you reverse your order of operations
to get x squared by itself.
So first, you add 9 to both sides then you divide both sides
by 4.
Next, you take the square root of both sides.
Of course you have two square roots
so you tack on the plus or minus.
So we have answers of plus or minus 3/2.
So anytime your equation does not have a linear term,
the easiest way to solve the equation would be
by taking square roots.
If your equation looks like this,
1x squared plus bx plus c equals 0 and b is even,
then completing the square is really fast.
For example, on this equation, it's in that form and 8 is even,
half of 8 is 4, square is 16 so I add 16 to both sides.
The left-hand side becomes the square of x minus 4.
I take the square root of both sides,
I add 4 to both sides, and I'm done.
That's really a lot quicker than using the quadratic formula.
So, try that method if you have 1x squared
and your linear term has a even coefficient.
If your equation looks like this,
where you have a perfect square, then you should solve
by taking square roots.
If we look at the last problem over here, our first goal was
to complete the square to get a perfect square.
Well on our problem, we already have the perfect square.
We just need to get rid of the 3 and the negative 4 and the 8.
We just need to get rid of these numbers.
OK. So we add 4 to both sides.
We divide both sides by 3.
We get the numbers out of the way.
And now, we're on the same step
that we were here on the last problem.
Now we take the square root of both sides.
One difference between the last problem
and this problem is what's
under the radical this time is a perfect square.
So this time, instead of having the square root of 19
which we couldn't do anything with,
now we have the square root of 4 which is 2.
So now, we will be able to simplify our answer.
So now I've got negative 2, plus or minus 2,
so I have negative 2 plus 2 or negative 2 minus 2
and I actually can find out my answers in simplified form.
If you equation involves fractions or decimals,
multiply both sides of the equation
by the appropriate quantity to get rid
of the fractions or decimals.
To clear decimals, you would multiply
by the appropriate power of 10, right, 10 or 100 or 1000.
I personally don't clear decimals though.
If I have decimals
in my quadratic equation, I leave them.
And I simply use the quadratic formula for those problems.
For fractions, I do usually clear them.
But it's personal preference.
And what do I mean by clear?
I multiply by the LCD.
Least common denominator.
So multiply every term by the least common denominator
which clears the fractions.
And then look and see, does this equation factor?
OK, then use that method.
Would taking square roots be the easiest method?
Complete the square?
Quadratic formula?
If you're doing a word problem or any problem
where you're allowed to use an approximation, hen go straight
to the quadratic formula.
Substitute in your values for a, b, and c in the formula
and then immediately pull out your calculator
and do an approximation.
You don't have to do any simplification,
you don't have simplify the radical, you don't have
to do any factoring,
just immediately find an approximations
for the two answers, OK?
Here we're going to choose the easiest method.
We don't actually have to solve any of these equations.
So what's the easiest method on the first one?
We've got 1x squared and we have an even linear coefficient.
So that would be completing the square.
Yes? The next one, we have no linear term.
No linear term means solve by taking square roots.
The next one, we have no constant term
which means, solve by factoring.
The next one, we have a perfect square, right?
We have that binomial perfect square there which means solve
by taking the square roots.
The next one, we have a mess.
And we want to put that mess right
into the quadratic formula.
The next one, we have all three different terms
and it doesn't factor.
All three terms and it doesn't factor.
It definitely doesn't look like an easy one to do
by completing the square so we're going
to use the quadratic formula.
The last one, does factor.
Yeah? It does factor.
It doesn't look like an easy to complete the square on because
of the 5 being odd, right?
And it does factor.
So those would be the easiest methods for each of those.