>> Welcome to the Cypress College Math Review
on Applications of Quadratic Equations.
Objective one, perimeter and area problems.
Here's some formulas that you need to know.
For a square, the perimeter is 4 times the length of a side.
The area is side squared.
For a rectangle, the perimeter or distance around,
is 2 times the length plus 2 times the width.
The area is length times width.
For a triangle, the perimeter is the sum of the three sides,
A plus B plus C. The area is 1/2 base times height
where the height and base are perpendicular to each other.
For a circle, the distance around is called circumference.
And there's two different formulas you can use,
2 pi R or pi times D. And the reason
for that is the diameter is twice the radius.
The area of a circle is pi times the radius squared.
The area of a triangle is 112 square meters.
The height is 2 meters longer than the base.
What are the dimensions of the triangle?
So we're told that the area
of the triangle is 112 square meters.
The formula for area
of a triangle is 1/2 base times height.
So we have to figure out what to put
in for the base and for the height.
The height is 2 meters longer than the base.
So height equals 2 plus base, longer than, plus base.
So we would let X be the base.
Then the height would be 2 plus X. Substitute in X for base
and 2 plus X or X plus 2 for the height and we have our equation.
Pause the video if you'd like to go through the steps
to solve this equation.
Here's the answer.
The area of a circle is 12 square inches.
What is the radius of the circle?
So area of a circle equals 12.
Our formula is area equals pi R squared
or pi times the radius squared.
So our unknown is the radius.
We'll let R be the radius.
We substitute in 12 for the area and here's the equation
that we need to solve.
Pause the video if you want to go
through the steps to solve it.
And then here's our answer.
Objective two, work rate problems.
It takes Guiana 2 hours longer
to paint a room than it does Bryson.
It takes 7 hours for the two of them to paint the room
if they are working together.
How long does it take each one of them
if they did the job all by themselves?
So we have a work rate problem.
And the set up is the part of the job done by Guiana
in 1 hour plus the part of the job done by Bryson
in 1 hour equals the part of the job done together in 1 hour.
The formula is 1 over time.
Now, Guiana takes 2 hours longer to paint the room
than Bryson does, so we should let X be the time it would take
Bryson to paint the room if he was working by himself.
Then X plus 2, 2 hours longer, is the time it would take Guiana
to paint the room working alone.
So the part of the job done by Guiana in 1 hour is 1
over how long it would take her to do it
by herself, which is X plus 2.
The part of the job done by Bryson in 1 hour is 1
over the time it would take him to do it by himself,
so that's 1 over X. And the part of the job done together
in 1 hour is 1 over 7, since 7 hours is how long it would take
them if they were working together.
So this is the equation that models that problem.
Pause the video if you want to go through the steps
to solve this equation.
And here's our answer.
Pause the video and try these problems.
Objective three, Pythagorean Theorem applications.
If you have a right triangle, then the sum of the squares
of the legs is equal to the square of the hypotenuse.
Example, the diagonal of a square is 26 feet.
What is the length of one of the sides?
So we have a square.
Here's its diagonal.
So the diagonal is 26 feet.
Since this is a square, all the sides are the same.
We'll let X be the length of the side of the square
and we fill that in our diagram.
We'll use the Pythagorean Theorem to solve for X,
substituting in X for A and X for B.
And the hypotenuse is 26, that's the diagonal.
Here's the equation that models our problem.
Pause the video if you want to go through the steps to solve.
And here's our answer.
A guy wire for an antenna is 27 yards long.
The height of the antenna is 18 yards.
How far from the base
of the antenna should the guy wire be anchored to the ground?
Here's an antenna and here's the guy wire that anchors it.
So a guy wire is a diagonal from the top
of the antenna down to the ground.
That forms a right triangle.
The length of the guy wire was 27 yards long.
The height of the antenna is 18 yards.
We're trying to find out how far from the base
of the antenna the guy wire should be anchored
to the ground.
Well let X be the distance from the base of the antenna
to where the guy wire will be anchored.
Now, we have a right triangle with the three sides filled in.
Here's the Pythagorean Theorem that we're going to use.
We'll let X be A and 18 be B. Those are interchangeable,
but C has to be the 27.
Here's the equation that models our problem.
Pause the video if you want to go through the steps
to solve this equation.
And here's our answer.
Objective four, consecutive integer problems.
Here are some consecutive integers, 3, 4, 5.
If we let X be the smallest integer, 3,
how do we get from 3 to 4?
Well, we would add 1.
How do we get from 4 to 5?
We would add 1 again.
So if the consecutive integers were unknown, we could set
up the problem using X for the smallest, X plus 1 for the next,
and X plus 2 for the one after that and so on and so forth.
What about consecutive even integers, like 4, 6, 8?
If X was the smallest, how do I get from 4 to 6?
Well, you would add 2.
Then you would add 2 again to get from 6 to 8.
So if the first one is X, the next one would be X plus 2.
The next one would be X plus 4 and so on and so forth.
Consecutive odd integers.
This is the one that messes most people up.
Here's my example, 7, 9, and 11.
How do we get from 7 to 9?
We add 2. How do we get from 9 to 11?
We add 2. We don't add 1, we don't add 3.
A lot of people think it's going
to be something different for odd integers.
But to get to the next odd integer, you have to skip
over the even integer that's in between them.
So you're going to be adding 2 no matter whether you're working
with consecutive even or consecutive odd.
So for consecutive off integers, you're going to set it
up where X is the smallest, X plus 2 is the next one,
and then X plus 4 and so on and so forth.
The product of two consecutive odd integers is 23 more
than 5 times their sum.
Find the integers.
Since we have consecutive odd integers,
I'll represent them as X and X plus 2.
Now let's read the sentence again.
The product, so product, is, that's the equals, 23 more than,
so that would be 23 plus, 5 times,
so it would be 5 times something, their sum.
Sum. So that sentence translates into this equation.
Product equals 23 plus 5 times sum.
It's important for you to be able to translate
in between a sentence and an equation.
Now what do we write down for the product
since our numbers are X and X plus 2?
Well, we would multiply them.
So multiply X times the quantity X plus 2.
That's equal to 23 plus 5 times their sum.
What do we put in for sum?
That means add them.
So we're going to add X and X plus 2.
This is the equation that models our problem.
Pause the video if you want to go through the steps
to solve the equation.
Here's our answer.
Be sure you have both pairs of solutions.
It's a good idea to check your answer when you can.
Here's the check for this problem.
Pause the video and try these problems.
Objective five, projectile problems.
The height in feet of a ball thrown upward from a height
of 18 feet is given by H equals negative 16 T squared plus 32 T
plus 18 after T seconds.
So H is the height of the ball and T is the time.
How many seconds will it take for the ball
to reach a height of 30 feet?
We want the height to equal 30 feet.
What's the height equal to?
Well, the height's equal to that quadratic expression.
We're going to let T be the time since the ball was thrown.
We're going to replace H with 30 because we want to know
when is the height equal to 30.
Here's the equation that models our problem.
Pause the video if you want to go
through the steps to solve it.
Notice on our answer that we have two different times.
So on the way up, after a half a second,
the ball reached 30 feet.
And then on the way back down, after 1 and a half seconds,
the ball was at 30 feet once again.
The height in feet of a ball thrown upwards from a height
of a6 feet is given by H equals negative 16 T squared plus 32 T
plus 6 after T seconds.
So the height is H and the time is T. How many seconds will it
take for the ball to hit the ground?
Well, if it's on the ground what's the height?
The height would be zero.
So we're going to let the height equal zero.
Well, the height is equal to this quadratic expression.
We're going to let T be the time since the ball was thrown.
We're going to let H be zero.
And solve for T. So here's the equation
that will give us the time when this ball hits the ground.
Pause the video if you'd like to go
through the steps to solve it.
And here's our answer.
Objective six, compound interest problems.
For interest that is compounded annually,
annually means one time per year,
use the formula A equals P times the quantity 1 plus R raised
to the T power.
Where A is the amount in the account at the end of the term,
P is the amount invested or the principal, T is the number
of years, and R is the interest rate.
Find the rate R at which $500 compounded annually grows
to $525 in two years.
Round your answer to the nearest tenth of a percent.
Since the money's being compounded annually,
we use this formula and we don't know the interest rate,
so let R be the unknown interest rate.
$500 is growing to $525, which tells us
that 500 is our principal and $525 is the amount we have
at the end, which is A. Two years, so T is equal to 2.
Substitute these into our formula and here's the equation
that models that problem.
The first step to solve this is to get
that quantity squared by itself.
We want to get rid of the 500.
It's being multiplied times that parentheses,
so we would divide both sides by 500.
Then you take the square root of both sides
and we have 1 plus R. Then we subtract 1 from both sides.
One of the answers will be negative.
We discard that one.
And the other answer is listed here.
The question asked us to round our answer
to the nearest tenth of a percent.
That can be tricky for people.
I recommend you first change it into a percent, like I did here,
then round to the nearest tenth of a percent.
So here's our answer.
Objective seven, distance, rate, time problems.
Sharon and Jill each drove from their home in St. George,
Utah to Salt Lake City, a distance of 330 miles.
Sharon drove 11 miles an hour faster than Jill did.
Sharon arrived 1 hour before Jill did.
How fast did each of them drive?
All right.
What are we given here?
Well, we're given the distance.
Each of them drove 330 miles.
It's only listed once, but that's the distance
that each of them drove.
So we are given both of their distances.
Any time you're given both of the distances, the best way
to set it up is using the time.
So what relationship do we know
about the times for Sharon and Jill?
Sharon arrived 1 hour before Jill did.
So her time was 1 hour less.
So the time for Sharon is 1 hour less than the time for Jill.
One hour less than, that reverses the order,
so it would be time for Jill minus 1.
So the time for Sharon is 1 hour less than the time for Jill.
Formula for distance.
Distance equals rate times time.
So time, which is what we need, is distance over rate.
So we substitute distance over rate for time for Sharon
and for the time for Jill.
What are our unknowns?
We don't know their rates.
We know that Sharon drove 11 miles an hour faster
than Jill did.
So we have Sharon's rate in terms of Jill's rate,
which means X should be Jill's rate.
If we let X be Jill's rate,
then Sharon's rate would be X plus 11.
Now we're ready to substitute in to our equation.
The distances are both 330 miles.
Sharon's rate was X plus 11.
Jill's rate was X. Here's the equation
that models our problem.
Pause the video if you want to go
through the steps to solve it.
And then here's our answer.
Pause the video and try these problems.