>> Welcome to the Cypress College Math Review on solving quadratic equations.
There are four methods for solving quadratic equations.
Solve by taking square roots, which is also called extracting square roots.
Solve by factoring using the zero product rule.
Solve by completing the square.
And solve by using the quadratic formula.
We are going to review each of these different methods.
Objective one, solving by taking square roots.
Let's review this method.
First of all, in this equation we notice that the only X is in a square.
This equation will solve very easily using this method.
You want to isolate the perfect square.
The perfect square is the X squared.
Notice that there are no other Xs in the equation.
So first we add 3 to both sides.
Then we divide by 2.
Now we have the perfect square completely isolated.
The next step is to take the square root of both sides.
Now on the left-hand side you get the absolute value of X which will give you plus or minus X.
And when you solve for X you get plus or minus the square root of 5.
No one actually shows those steps.
Most people go straight from this equation, X squared equals 5,
to X equals plus or minus the square root of 5.
For this problem the perfect square is the square of X plus 4.
Notice that there are no other Xs in the equation.
So our first step is to isolate that.
So we add 5 to both sides.
Then divide both sides by 2.
Now we take the square root of both sides.
So X plus 4 is equal to plus or minus the square root of 12.
We need to take out any perfect squares that we can out of
that radical, and 4 is a perfect square.
So we have our answer of X equals negative 4 plus or minus 2 times the square root of 3.
Pause the video and try these problems.
Objective two.
Solve by factoring.
In this equation we not only have the quadratic term 5X squared,
but we also have the linear term negative 14X.
So X is not just in that perfect square X squared.
So our first method won't work.
To solve by factoring, you need to move all terms to one side.
I always like to have a positive leading coefficient
when I'm factoring so I make sure that happens.
I get a positive 5X squared.
And the right-hand side you have to have a zero.
It has to equal zero.
The next step is to factor your trinomial.
To get 5X squared I would have 5X and X. There's different options for 8.
I could have 1 times 8 or 2 times 4.
It will work out that I need a 4 here and a 2 there.
This is 4X and this is 10X.
Those would add to get 14X.
The signs need to be the same because this is a positive.
And they both need to be plus to be able to make it so that I have positive 14X.
Now I have it factored.
By the zero product rule we know that if A times B is equal to zero,
then either A is equal to zero or B is equal to zero.
So I have the quantity 5X plus 4 times the quantity X plus 2 is equal to 0.
That means that either the first factor is equal to 0 or the second factor is equal to 0.
Now we solve each of these individual equations.
So we have X equals negative 4/5 or X equals negative 2,
giving us the two answers to this equation.
Our next example we will solve by factoring again.
We want a positive leading coefficient so we move everything to the left.
Equals 0. Now we factor.
So we have 5X and 3X and we need a 2 here to give us 6X.
And a 4 here to give us 20X.
The signs are going to be different because the 8 is negative.
And I'm going to need a negative 20X and a positive 6X to get the negative 14X that I need.
Now the left-hand side is factored.
Now we're going to use the zero product rule.
Since A times B is equal to 0, either the first factor is equal to 0
or the second factor is equal to 0.
Then we solve each of these equations.
So we get X equals negative 2/5 or X equals 4/3 which are the two solutions to this equation.
Pause the video and try these problems.
Objective three.
Solve by completing the square.
To solve by completing the square, we first move the constant term to the other side.
So we add 3 to both sides for this equation.
And then we divide by the leading coefficient so that you have a 1X squared.
In this equation we already have 1X squared so we don't need to do that.
Next you complete the square.
To complete the square you take half of the negative 10 which is negative 5,
square it, and add it to both sides.
So to complete the square we have to add 25 to both sides.
What you do to one side you have to do to the other.
You're always adding when you complete the square.
Next you write the left-hand side as a perfect square.
X squared.
This would be an X. This is minus so this is minus.
And half of 10 was 5.
Check it out.
If you square X minus 5 you do get X squared minus 10X plus 25.
The right-hand side is 28.
Now we use our first method.
We take the square root of both sides.
So X minus 5 is equal to plus or minus the square root of 28.
Now we simplify that radical and that's 4 times 7.
So the right-hand side becomes plus or minus 2 times the square root of 7.
We solve for X by adding 5 to both sides.
So we get 5 plus or minus 2 times the square root of 7.
Sometimes it's okay to write your answers like that.
Other times your instructor wants them written separately.
So 5 plus 2 times the square root of 7 or 5 minus 2 times the square root of 7.
Those are the solutions to this equation.
Let's solve this equation by completing the square.
So the first step is to move the constant term to the other side.
So we add 8 to both sides.
The next step is to divide by the leading coefficient so that we have a 1X squared.
Be sure you divide every single term on both sides of the equation by that coefficient.
So we have X squared plus 6X is equal to 2.
Now we complete the square.
Half of 6 is 3, squared is 9.
So we add 9 to both sides.
The left-hand side becomes a perfect square.
X squared becomes X plus, plus, half of 6 was 3.
The right-hand side is 11.
Now we take the square root of both sides.
So X plus 3 equals plus or minus the square root of 11,
and the square root of 11 cannot be simplified.
So our answers are negative 3 plus or minus the square root of 11.
Pause the video and try these problems.
Objective four.
Solve by using the quadratic formula.
So for this equation we're going to solve using that method.
So we first move all terms to one side and get them organized.
So we add 12X to both sides.
We have to have it equal to 0.
So we must have it in the form AX squared plus BX plus C equals 0.
From that we can determine that A is 3, B is 12, and C is 7.
Next we write down the formula.
So X equals negative B plus or minus the square root of B squared minus 4AC all over 2A.
Use parentheses when you're replacing variables with numbers.
So I have parentheses everywhere I see a letter.
Now I substitute in.
So B is 12, A is 3, and C is 7.
Now we simplify this.
So we have X equals negative 12 plus or minus the square root of -- And 12 squared is 144.
And then minus 84.
Then that's all over 6.
So we get negative 12 plus or minus the square root of 60 all over 6.
Be careful not to cancel the 12 and the 6.
You can only cancel factors, and the numerator is not factored.
We have to simplify that radical.
60 is 4 times 15.
So we have 2 times the square root of 15.
Now we can factor out a 2.
We do so and we end up with negative 6 plus or minus the square root of 15 all over 6.
Now we can cancel.
2 goes in to here once and 2 goes in to here 3 times.
So our answers to this equation are negative 6 plus or minus the square root of 15 all over 3.
We're going to solve this using the quadratic formula again.
So we move all terms to one side.
Have it equal to 0.
So we have it in the form AX squared plus BX plus C equals 0.
That tells us that A is 5, B is negative 8, and C is negative 3.
Next we write the formula.
X equals negative B plus or minus the square root of B squared minus 4AC all over 2A.
Then we use parentheses since we're replacing variables with numbers.
So we have that B is negative 8, A is 5, and C is negative 3.
Now we simplify this.
So we have positive 8 plus or minus the square root of 64
because that's negative 8 quantity squared, and then watch out for your signs.
We've got two negatives so we end up with plus 60 all over 10.
So we have X equals 8 plus or minus the square root of 124 all over 10.
124 is 2 times 62 and 62 is 2 times 31.
So X equals 8 plus or minus the square root of 4
because we have 2 squared times the 31 all over the 10.
I'm trying to simplify the radical.
So we get 8 plus or minus 2 rad 31 all over 10.
Now we factor out a 2 and we're left with 4 plus or minus the square root of 31 all over 10.
Now you can cancel.
So many people coming back over here will cancel with the 8 and the 10.
And you cannot cancel the 8 with the 10 because the 8 is not a factor of the numerator.
Looking over here on the right, the 2 is a factor of the numerator.
Since 2 is a factor of the numerator, I can cancel and I get 1 over 5.
So the solution to this equation is 4 plus or minus the square root of 31 all over 5.
Pause the video and try these problems.
Objective five.
Determine the best method to solve a quadratic equation.
In the equation AX squared plus BX plus C equals 0, AX squared is a quadratic term,
BX is called the linear term, and C is the constant.
If your equation does not have a constant term, then solve by factoring.
That's definitely going to be the easiest method.
In this equation we move all terms to one side and we notice that there's no constant term.
Every single term has an X in it.
So we factor, set each factor equal 0, and then solve.
Any time the equation factors easily then that's the easiest method to solve.
Solve by factoring if the equation factors easily.
If your equation does not have a linear term, then solve by taking square roots.
The only place you see the X is in the perfect square X squared.
So you're going to solve by taking square roots.
So in this equation we move 9 to the other side, divide by 4,
and take the square root of both sides.
So our answers are X equals plus or minus 3/2.
If your equation looks like 1X squared plus BX plus C equals 0, and B is even,
then completing the square is really fast.
Any time you have a 1X squared and you have an even linear coefficient, B is even,
then you should complete the square.
So for this equation we have 1X squared minus 8X, and then we have a constant term.
The constant term can be on either side, but definitely for this equation solve
by completing the square is the easiest.
So we have the constant on the other side.
That's the first step, already done for us.
Now we complete the square.
Half of 8 is 4, squared is 16.
We add 16 to both sides.
The left-hand side we write as a perfect square.
Then we take the square root of both sides of the equation.
So we have X minus 4 equals plus or minus the square root of 19.
We add 4 to both sides and we get our solutions, 4 plus or minus the square root of 19.
If your equation looks like this where the only place you see the X is in a perfect square,
then solving by taking square roots is the easiest method.
So we want to get the perfect square by itself by adding 4
to both sides and then dividing by 3.
Then we take the square root of both sides.
So we get X plus 2 equals plus or minus the square root of 4, but the square root of 4 is 2
so now we have to write it as two separate equations because our radical disappeared.
We subtract 2 from both sides and we get X equals negative 2 plus 2 which is 0
or X equals negative 2 minus 2 which is negative 4,
and those are the two solutions to this equation.
If your equation involves fractions, then multiply both sides of the equation
by the appropriate quantity, which of course is the least common denominator,
to get rid of the fractions.
Then use whatever method is easiest to solve the resulting equation.
If your equation involves decimals,
then the quadratic formula is probably the easiest method.
If you're working on a word problem or any problem where you're allowed
to use an approximation, then the quadratic formula is usually the fastest.
Substitute A, B, and C in to the formula,
then pull out a calculator and approximate the answers.
No need to simplify.
Pause the video and try these problems.
Objective six.
Practice solving by any method.
For each of the problems in the exercise set, choose the easiest method,
then solve the quadratic equations.
Pause the video and try these problems.