QUADRATIC FUNCTIONS: Problem Set 1
NOTES:
The motion of projectiles can be represented by a quadratic equation. Here are two examples of this kind of motion:
- A Ball being thrown upwards into the air.
- An aircraft dropping a watermelon.
***
You are going to plot a graph to show the motion of a cannonball being fired into the air at an angle of 45 degrees.
The equation for the motion is:
2
y = ax + bx
Where a = -1/500 and b = 1.
2
In the equation y = (-1/500)x + x, y represents the distance in meters and x is the time in seconds.
The equation describing the motion of a projectile is a quadratic equation.
The graph of a quadratic equation is a parabola.
The value of y rises to a maximum and then falls back to zero.
WORKSPACE:
ANS:
45 , 80 , 105 , 120 , 125 , 120 , 105 , 80 , 45 , 0
SOLVING QUADRATIC EQUATIONS: Factoring
- Solve quadratic equations by factoring and using the zero-product property
NOTES:
In past activities we factored and graphed equations of the form:
2
y = ax + bx + c
This activity focuses on the particular case when y = 0. That is, we will learn to solve equations of the form:
2
0 = ax + bx + c
To do this, we will use factoring.
***
Let's see how to graph the following equations:
2
y = x - 5x - 14
2
x|y y = (-3) - 5(-3) - 14
--+-- 9 + 15 - 14 = 10
-3|10
-2|0
0|-14
***
Before pressing ENTER, describe in your own words the points where the graph crosses the x-axis.
x|y
--+--
-3|4
-2|0
0|-2
1|0
The points where the graph crosses the x-axis are the points where y = 0.
***
Before pressing ENTER, describe in your own words the solutions to equations like 0 = x^2 - 7x + 12.
The solutions to the equation are the x-values of the points where the graph crosses the x-axis.
***
It is not always practical to graph an equation to determine the solutions when y = 0.
In the next part of the lesson, algebraic procedures will be used to find solutions to equations like:
2
ax + bx + c = 0
Without graphing.
First, let's review a few facts that will be needed.
***
Complete the folling products:
3 * _ = 0
_ * 5 = 0
? * ? = 0
Notice that the last product is correct for any pair of numbers as long as one of the numbers is zero.
***
Before pressing ENTER, describe in your own words when:
a * b = 0
Either a or b (or both) must be zero. This is known as the zero-product property.
It also will be true that:
(x - 3)(x + 5) = 0
If either:
x - 3 = 0 OR x + 5 = 0
So x = 3 or x = -5.
***
Before pressing ENTER, describe in your own words when:
(ax + b)(cx + d) = 0
It will be true if either:
ax + b = 0 or cx + d = 0
***
Notice what happens when we factor the following equation:
2
Start with... 3x - x = 0
Then factor... x(3x - 1) = 0
Then either... x = 0 OR 3x - 1 = 0
So either... x = 0 OR x = 1/3
Here are some more examples of solving equations:
2
4x - 9 = 0
(2x + 3)(2x - 3) = 0
x = -3/2 OR 3/2
And:
3 2
3x - 6x - 24x = 0
2
3x(x - 2x - 8) = 0
3x(x - 4)(x + 2) = 0
x = 0, 4 or -2
***
Before pressing ENTER, describe in your own words a way to solve equations of the form:
2
ax + bx + c = 0
Factor the equation, set each factor equal to zero and solve the resulting equations.
WORKSPACE:
2
8x + 5x = 0
x(8x + 5) = 0
0 OR -5/8
***
2
x + 8x + 16 = 0
-4
***
2
x - 2x - 15 = 0
-3, 5
ANS:
0 , -14 , -18 , -18 , -14 , 0 , 4 , 0 , -2 , -2 , 0 , 4 , 10 , 0 , 0 , zero , 0 , 0 ,-9 , 5/2 , 0 , 0 , 0 , 4/3 , -3/2 , 0 -2 , -1/2 3 , 6/5 -4 , 7/3 -9/2 , 0 -2 5 , -3/2 5/4 , 0 -5/8 , -5 5 , -4 , 6 , -3 5 , 1 -1 , 3/2 -9 , ... , 7 3 , -9/5 -1 , 1/3 7 , -4 5 , 1/4 -1/4 , 8 2 , 10/3 -10/3
SOLVING QUADRATIC EQUATIONS: Completing the Square
2
- Solve equations of the form a(x + b) = c.
- Solve quadratic equations by completing the square.
NOTES:
In the last activity we solved quadratic equations by factoring.
2
x + 7x + 12 = 0
(x + 3)(x + 4) = 0
However, not all equations are easily factored. We need some new methods to help us solve quadratics.
The first method we will learn is called COMPLETING THE SQUARE.
***
Sometimes our solutions will involve irrational numbers like:
_ _ _
\/5 2\/3 3 + \/7
Let's see how we get these solutions!
Look at this equation:
2
x = 9
If we take the square root of both sides, we find the two solutions.
__
/ 2 _
\/ x = \/9
|X| = 3
x = 3, -3
Here are some other examples:
2
x = 7
_ _
x = \/7, -\/7
And:
2
4x = 25
2 25
x = --
4
5 -5
x = -, --
2 2
***
Before pressing ENTER, describe in your own words how we solve the following quadratic equation:
2
ax = b
Divide both sides by a and then take the square root of each side:
2
ax = b
2 b
x = -
a
_ _
/b /b
/ - , - / -
x = \/ a \/ a
***
We can use this method to solve the following quadratic equations:
2
If, (x -4) = 36
Then, (x - 4) = 6 OR -6
Here are some more examples:
2
3(x - 6) = 75
2
(x - 6) = 25
(x - 6) = 5 OR -5
Here are some more examples:
2
(x - 2) = 5
_ _
(x - 2) = \/5 OR -\/5
_ _
x - 2 = \/5 OR x - 2 = -\/5
_ _
x = 2 + \/5 OR x = 2 - \/5
***
Before pressing ENTER, use your own words to describe how we solve the following quadratic equation:
2
a(x + b) = c
First we divide by a. Next take the square root of both sides. Then we solve the resulting equations.
2
a(x + b) = c
2 c
(x + b) = -
a
_ _
/c /c
/ - , - / -
(x + b) = \/ a \/ a
_ _
/c /c
/ - OR / -
x = -b + \/ a x = -b - \/ a
The solution steps we have been using don't always work well because some equations like:
2
ax + bx + c = 0
Are not easy to factor.
To solve any equations of this form we will use two facts about PERFECT SQUARE TRINOMIALS.
2 2 2
(x + b) = x + 2bx + b
2 2 2
(x - b) = x - 2bx + b
Notice how we solve the following equation:
2
x + 6x + 7 = 0
Since we can't factor this equation let's change it to a Perfect Square Trinomial which we can factor...
2
x + 6x + 7 = 0
[ 2
/ x + 6x = -7
|
\ 2
[ x + 6x + 9 = -7 + 9
2
(x + 3) = 2
_ _
(x + 3) = \/2 OR -\/2
_ _
x + 3 = \/2 OR x + 3 = -\/2
_ _
x = -3 + \/2 OR x = -3 - \/2
The procedure in brackets is a process called Completing the Square.
Here are some more examples of completing the square:
2
x - 8x + 4 = 0
2
x - 8x = -4
2
x - 8x + 16 = -4 + 16
2
(x - 4) = 12
_ _
x - 4 = 2\/3 OR -2\/3
_
x = 4 +/- 2\/3
And:
2
x + 7x + 2 = 0
2
x + 7x = -2
2 49 -8 49
x + 7x + -- = -- + --
4 4 4
2 41
(x + 7/2) = --
4
WORKSPACE:
2
16x = 9
x = 3/4, -3/4
"This is correct because 16(3/4)(3/4) = 9 and 16(-3/4)(-3/4) = 9.
***
ANS:
4 -4 , 10 -10 , 3\/2 -3\/2 , 2/3 -2/3 , 3/4 -3/4 , 5/7 -5/7 , \/3 -\/3 , 10 , -2 , 11 , 1 , 12 -2 , -6 + \/11 -6 - \/11 , -2 + \/3 -2 - \/3 , 5 -4 , 3 + \/7 3 - \/7 , 5 + \/7 5 - \/7 , 7/3 -5 , 1 + \/3 1 - \/3
The above is continued. Current Date: February 28, 2006
NOTES:
Here are some more examples of completing the square:
2
x - 8x + 4 = 0
2
x - 8x = -4
2
x - 8x + 16 = -4 + 16
2
(x - 4) = 12
_ _
x - 4 = 2\/3 OR -2 \/3
_
x = 4 +/- 2\/3
And:
2
x + 7x + 2 = 0
2
x + 7x = -2
2 49 -8 49
x + 7x + -- = -- + --
4 4 4
2 41
(x + 7/2) = --
4
__ __
\/41 \/41
x + 7/2 = ---- OR - ----
2 2
__
-7 +/- \/41
x = -----------
2
***
Before pressing ENTER, describe in your own words one way to solve polynomials of this form:
2
x + bx + c = 0
Complete the square on the left side:
2 2
2 b b
x + bx + -- = -c + --
4 4
factor,
2
b 2 b
(x + -) = -c + --
2 4
Then take the square root of both sides and solve.
***
Let's see how the transformation for solving an equation by completing the square affect the graph of the equation.
2
x - (3/2)x - 1 = 0
2
x - (3/2)x = 1
2
x - (3/2)x + 9/16 = 1 + 9/16
2
(x - 3/4) = 25/16
x - 3/4 = +/- 5/4
x - 3/4 = 5/4 OR x - 3/4 = - 5/4
x = 2 x = -1/2
Here is another graphic example of solving equations:
2
x + 2x - 1 = 0
2
x + 2x = 1
2
x + 2x + 1 = 1 + 1
2
(x + 1) = 2
_
x + 1 = +/- \/2
_ _
x + 1 = \/2 OR x + 1 = -\/2
_ _
x = -1 + \/2 x = -1 - \/2
~ ~
x = .41 x = -2.41
***
Before pressing ENTER, describe in your own words how transformations of equations affect the solutions to the equations.
As we have seen, the solutions are not affected, but always lie on the same lines.
WORKSPACE:
2
x - 4x + 1 = 0
2
x - 4x = -1
2
x - 4x + 4 = -1 + 4
2
(x - 2) = 3
_ _
x - 2 = \/3 OR \/3
_ _
x = 2 + \/3, 2 - \/3
*Why do they pick that certain number for completing the square? I'm not sure... I'll have to figure that out*
***
2
x + 2x - 1 = 0
2
x + 2x = 1
2
x + 2x + 1 = 1 + 1
2
(x + 1) = 2
_ _
x + 1 = \/2 OR -\/2
_
x = -1 + \/2
***
2
x + 3x - 8 = 0
2
x + 3x = -8
2
x + 3x + 9/4 = 41/4
2
(x + 3/2) = 41/4
__ __
x + 3/2 = \/41/2 OR -\/41/2
__ __
x = (-3 + \/41)/2, (-3 - \/41)/2
ANS:
2 + \/3 2 - \/3 , -1 + \/2 -1 - \/2 , (-3 + \/41)/2, (-3 - \/41)/2 , 9 -9 , 3/7 -3/7 , 2\/2 -2\/2 , 11 3 , -2 -8 , 9/2 -3/2 , -3 + \/2 -3 - \/2 , 4 + \/7 4 - \/7 , (-3 + \/33)/2 (-3 - \/33)/2
SOLVING QUADRATIC EQUATIONS: The Quadratic Formula
2
- Solve equations of the form ax + bx + c = 0 using the quadratic formula.
NOTES:
In the last activity we learned the method of completing the square which can use to solve all quadratic equations of the form:
2
ax + bx + c = 0
In this activity we will learn a "shortcut" for solving these same quadratic equations. This shortcut is the Quadratic Formula.
***
Here is a quadratic equation:
2
3x + 2x + 1 = 0
Written in the form:
2
ax + bx + c = 0
In the above equation,
a = 3
b = 2
c = 1
***
2
ax + bx + c = 0
Determine the coefficients a, b, and c in the following quadratic equation:
2
4x + 6x + 2 = 0 a = 4, b = 6, c = 2
***
Using algebraic transformations, and completing the square, we can solve the following quadratic equation for x:
2 /
ax + bx + c = 0 a = 0
The result is:
_______
/ 2
-b +/- \/ b - 4ac /
x = ----------------- a = 0
2a
This is a formula for finding the solution of all quadratic equations. It is called the quadratic formula. We will derive this formula in Algebra II.
***
_______
/ 2
-b +/- \/ b - 4ac
x = -----------------
2a
The way we read the quadratic formula is:
"x equals the opposite of b plus or minus the square root of b squared minus 4ac, all divided by 2a."
***
_______
/ 2
-b +/- \/ b - 4ac
x = -----------------
2a
Let's see how the quadratic formula works.
2
x - 3x + 2 = 0
a = 1
b = -3
c = 2
_____________
/ 2
-(-3) +/- \/(-3) - 4(1)(2)
x = --------------------------
2(1)
_
3 +/- \/1
= ---------
2
3 +/- 1
= -------
2
So,
3 + 1 3 - 1
x = ----- OR ----- = 2 OR 1
2 2
***
Let's solve the same equation by completing the square.
2
x - 3x + 2 = 0
2
x - 3x = -2
2
x - 3x + (9/4) = -2 + (9/4)
2
[x - (3/2)] = 1/4
x - (3/2) = +/- 1/2
x - (3/2) = 1/2 OR x - (3/2) = -1/2
x = 4/2 = 2 x = 2/2 = 1
Notice that we get the same answers.
*Huh? They look different to me...*
***
Let's do an equation together...
_______
/ 2
-b +/- \/ b - 4ac
x = -----------------
2a
2
2x + 6x + 4 = 0
a = 2
b = 6
c = 4
____________
/ 2
-(6) +/- \/(6) - 4(2)(4)
x = ------------------------
2(2)
_
-6 +/- \/4
x = ----------
4
-6 + 2 -6 - 2
x = ------ OR ------
4 4
x = -1 OR -2
***
Before pressing ENTER, use your own words to describe how we use the quadratic formula to solve equations like:
2
x - 5x + 9 = 0
Substitute 1, -5, and 9 for a, b, c in the formula, then simplify.
_______
/ 2
-b +/- \/ b - 4ac
x = -----------------
2a
***
WORKSPACE:
a = 1
b = 6
c = -16
_______
/ 2
-6 +/- \/ 6 - 4(-16)
x = --------------------
2
___
-6 +/- \/100
------------
2
-6 +/- 10
---------
2
2 OR -8
***
a = 1
b = 7
c = -3
______
/ 2
-7 +/- \/ 7 + 12
x = ----------------
2
__
-7 +/- \/61
-----------
2
***
a = 3
b = 35
c = 50
________
/ 2
-35 +/- \/ 35 - 600
x = -------------------
6
35 +/- 25
---------
6
***
a = 5
b = 1
c = -18
_______
/ 2
-1 +/- \/ 1 + 360
x = -----------------
10
-1 +/- 19
---------
10
9/5 OR -2
***
a = 1
b = 11
c = 24
_______
/ 2
11 +/- \/ 11 - 96
x = -----------------
2
11 +/- 5
--------
2
3 OR 8
***
a = 6
b = 17
c = 7
________
/ 2
-17 +/- \/ 17 - 168
x = -------------------
12
-17 +/- 11
----------
12
-1/2 OR -7/3
***
a = 5
b = 29
c = -6
_______
/ 2
-b +/- \/ b - 4ac
x = -----------------
2a
ANS:
4 6 2 , 3 -2 5 , 2 1/2 3 , 1 -4 -7 , 1 2 -4 , 2 6 4 , 6 , 6 , 2 , 4 , 2 , 4 , 2 , 2 , -1 , -2 , 1 -6 , 2 -8 , -1/2 1/3 , -3 + \/5 -3 - \/5 , 5/2 -7/3 , (-7 + \/61)/2 (-7 - \/61)/2 , 5/3 10 , 9/5 -2 , 3 8 , -1/2 -7/3
The above is continued. Current Date: March 6, 2006
NOTES:
This is the format:
2
ax + bx + c = 0
_______
/ 2
-b +/- \/ b - 4ac
x = -----------------
2a
WORKSPACE:
a = 5
b = 29
c = -6
________
/ 2
-29 +/- \/ 29 + 120
x = -------------------
10
-29 +/- 31
----------
10
***
a = 2
b = 19
c = 35
________
/ 2
-19 +/- \/ 19 - 280
x = -------------------
4
-19 +/- 9
---------
4
***
a = 1
b = -7
c = 12
_______
/ 2
7 +/- \/ -7 - 48
x = ----------------
2
7 +/- 1
-------
2
***
a = 28
b = -1
c = -15
_________
/ 2
1 +/- \/ -1 + 1680
x = -------------------
56
1 +/- 41
--------
56
***
a = 9
b = -22
c = 8
_________
/ 2
22 +/- \/ -22 - 288
x = -------------------
18
22 +/- 14
---------
18
***
a = 8
b = 25
c = 3
________
/ 2
-25 +/- \/ 25 - 96
x = ------------------
16
-25 +/- 23
----------
16
***
ANS:
1/5 -6 , -5/2 -7 , 4 3 , -5/7 3/4 , 2 4/9 , -1/8 -3
QUADRATIC FUNCTIONS: Problem Set 2
NOTES:
In Problem Set 1 we graphed points belonging to the quadratic equations that described the motion of a cannonball fired into the air.
2
y = (-1/500)x + x
In this problem imagine that the cannonball is being fired on Jupiter where the gravity is about 2.5 times that of the Earth.
As you might expect, gravity affects the flight of the cannonball.
The equations for the motion of the cannonball on Jupiter is:
2
y = (-1/200)x + x
Plot some points for this equation and later we'll compare the graphs for the effects of gravity.
WORKSPACE:
ANS:
0 , 22 , 38 , 47 , 50 , 47 , 38 , 22 , 0
QUADRATIC FUNCTIONS: The Distance Formula
- Excursion
NOTES:
Suppose one day you find an old map that looks like this:
--+----------+
R | |
| a b |
I| c |
| |
V | d |
| e |
E| |
|f |
R | g |
-----R---A----
-------O---D--
Written on the map is the following description:
"In a field 10 miles by 10 miles square,
Are found 7 trees, A-B-C-D-E-F-G
If you please.
If you're smart,
Find two trees 5 miles apart.
Halfway between these two
Is a treasure buried for you."
Between which two trees will you find the buried treasure?
At the end of this lesson we will return to this problem and you will be able to find the buried treasure!
***
In a previous lesson you learned about the PYTHAGOREAN THEOREM and some of its applications.
One of the most important applications of the Pythagorean Theorem is finding the DISTANCE between any two points in a plane.
The distance between any two points is always a positive number.
For example, to find the distance between these two points on the number line, you must compute the absolute value of the difference in their coordinates.
The distance between two points will always be the same.
This is true whether you go from point 1 to point 2 or from point 2 to point 1.
***
If two points are on the same horizontal line, the distance between them is the absolute value of the difference between their x-coordinates.
D = |10 - 2| = 8
If two points are on the same vertical line, the distance between them is the absolute value of the difference between their y-coordinates.
D = |5 - 2| = 3
***
But what if two points are NOT on the same vertical or horizontal line?
Now we have to find another way to determine the distance between the two points.
Let's add a VERTICAL line to the graph.
Now add a HORIZONTAL line to the graph.
You have made a RIGHT triangle.
The process of finding the lengths of the vertical and horizontal sides of a right triangle is similar to the process of finding the slope of a line, a skill that you learned earlier.
To calculate the slope of a line, you found the change in y-coordinates and divided it by the change in x-coordinates.
You can use the same information here to find the length of the third side of the triangle.
Tx = x - x Ty = y - y
2 1 2 1
a = |Tx| b = |Tx|
a - |x - x| b = |y - y|
2 1 2 1
Let side a equal 3 and side b equal 6, then solve for c:
2 2 2
c = a + b
2 2 2
c = 3 + 6
2
c = 9 + 36
2
c = 45
__ _
c = \/45 OR 3\/5
[This is the distance from (2,5) to (8,2)]
***
Let's review the procedure for finding the distance between any two points.
We will call these points point 1 (x^1,y^1) and point 2 (x^2,y^2) (Do not square, these are just labels!).
But remember that it doesn't matter which point is chosen as point 1 or point 2.
- First, make a right triangle.
- Measure length of the vertical and horizontal sides.
Now you can use the Pythagorean Theorem to find the length of the hypotenuse of this triangle.
Let a = |x - x| and b = |y - y|
2 1 2 1
Then solve for c:
2 2 2
c = |x - x| + |y - y|
2 1 2 1
Since any number squared is positive, the absolute value signs are not needed.
2 2 2
c = (x - x) + (y - y)
2 1 2 1
Finally, this is the result when we solve for c.
_________________
/ 2 2
c = \/(x - x) + (y - y)
2 1 2 1
This is the length of the third side of the triangle (the hypotenuse).
It is also the distance from (x^1,y^1) to (x^2,y^2).
_________________
/ 2 2
c = \/(x - x) + (y - y)
2 1 2 1
If we let D represent the distance between any two points (x^1,y^1) and (x^2,y^2) on a coordinate plane then we have the following equation:
_________________
/ 2 2
D = \/(x - x) + (y - y)
2 1 2 1
This equation is known as the Distance Formula.
The distance is the same no matter which point you start with.
***
To find the distance from the point (4,-2) to (-5,3), let:
x = 4 x = -5
1 2
y = -2 y = 3
1 2
Use these values in the Distance Formula.
_________________
/ 2 2
D = \/(x - x) + (y - y)
2 1 2 1
____________________
/ 2 2
D = \/(-5 - 4) + (3- (-2))
__________
/ 2 2
D = \/(-9) + (5)
_______
D = \/81 + 25
___
D = \/106
***
At the beginning of the lesson you were examining an old map.
To locate the buried treasure, you must find two trees that are 5 miles apart.
Tree A. (4,8)
Tree B. (8,8)
Tree C. (2,7)
Tree D. (5,5)
Tree E. (9,4)
Tree F. (1,2)
Tree G. (7,1)
Now you can use the Distance Formula to find the distance between any two trees.
Which two trees are 5 miles apart? Tree D and F.
X marks the spot halfway between the two trees where treasure is buried. HAPPY DIGGING!!!
*Editor's Note: Hehe, :-) ... O_o ... ;-)*
WORKSPACE:
ANS:
11 , 8 , 12 , 7 , 10 , 7 , 3 , 6 , \/40 , #3 , #1 , 5 , 13 , 10 , 4,8 , 8,8 , D F
QUADRATIC EQUATIONS AND THE DISTANCE FORMULA II: Test
This lesson will evaluate your skills in the solutions to quadratic equations and using the distance formula.
NOTES:
WORKSPACE:
ANS:
a , a , NOT:a,d , b , d , c , d , a , c , b
DIVINDING POLYNOMIALS: Long Division
- Divide polynomials using long division.
NOTES:
In our unit on polynomials we learned how to factor some polynomails by "guess and test."
In this topic we are going to investigate another method for determining factors of polynomials. That method is the method of long division.
We say that 13 is a factor of 32513 if 13 divides into 32,513 evenly.
32513/13 = 2501
Since the remainder is zero, 13 is a factor of 32,513.
We can use the method of long division to determine if 3K + 1 is a factor of 3k^2 + 4k + 1. Notice how it is similar to dividing 341 by 31.
1k +1
______________
3k + 1 )3k^2 + 4k + 1
-(3k^2 + k)
--------
3k + 1
-(3k + 1)
------
0
11
____
31 )341
-(31)
--
31
-(31)
--
0
***
Before pressing ENTER, describe in your own words one way to tell if "polynomial a" is a factor of "polynomial b."
Divide "polynomial b" by "polynomial a" using long division. If the remainder is zero, then "a" is a factor of "b."
***
WORKSPACE:
ANS:
k , 2k^2 + 5k , 4 , 8k + 20 , 3k^2 , 3k^3 + 12k^2 , -2k , -2k^2 - 8k , -5 , -5k - 20 , 2x , 10x^2 + 8x , -3 , -15x - 12 , x^2 , x^3 - 4x^2 , 2x^2 - 7x , 2x , 2x^2 - 8x , x - 4 , +1 , x - 4 , 0 , 3x^2 , 3x^3 + 6x^2 , -5x^2 - 3x , -5x , -5x^2 - 10x , 7x + 8 , +7 , 7x + 14 , -6 , 3x^2 , 6x^3 - 12x^2 , 12x^2 - 38x , +6x , 12x^2 - 24x , -14x + 28 , -7 , -14x + 28 , 0 , 3x^2 , 15x^3 - 6x^2 , -35x^2 - x , -7x , -35x^2 + 14x , -15x + 6 , -3 , -15x + 6 , 0 , 2x^2 , 6x^3 + 12x^2 , x^2 - 10x , +(1/3)x , x^2 + 2x , -12x - 4 , -4 , -12x - 24 , 20 , x^2 + 3x + 4 , 0 , 3x^3 - 2x - 6 , 7 , 5x^2 + 2x - 9 , 26 , 2x^2 + (1/2)x - 8 , 0 , 7x^3 + 5x^2 + 6x + 1 , 5
The above is continued (some parts are from this date, though). Current Date: March 14, 2006.
DIVIDING POLYNOMIALS: Exponents
- Simplify expressions of the form a^m / a^n.
- Review the meaning of zero and negative exponents.
NOTES:
In this topic we will learn how to simplify expressions like these:
7 3 3
a a a
-- -- --
3 7 3
a a a
Let's see how we simplify the quotient of expressions with the same base.
***
Write a general rule for the quotient of two expressions with the same base, where the exponent in the numerator is greater than the expeonent in the denominator.
m - n
***
Write a general rule for the quotient of two expressions with the same base, where the exponent in the denominator is greater than the exponent in the numerator (REVERSED).
n - m
***
Here is another rule for simplifying expressions like a^m / a^n where n > m.
We subtract the exponent in the denominator from the exponent in the numerator.
5
x 5-10 -5
--- = x = x
10
x
Negative exponents can be used to simplify expressions like x^5 / x^10.
***
8
x 8-14 -6
--- = x = x
14
x
Negative exponents can be used to simplify expressions like x^8 / x^14.
We know that:
5
x 1 1
--- = ----- = --
10 10-5 5
x x x
And:
5
x 5-10 -5
--- = x = x
10
x
So:
-5 1
x = --
5
x
***
Notice how we simplify the following expressions.
5 6
x y 2 5
---- = x y
3
x y
3 5 3
a b -1 3 b
---- = a b = --
4 2 a
a b
***
Before pressing ENTER, describe in your own words how we simplify expressions like:
7
a b
----
4 2
a b
Subtract the exponents of corresponding variables and then simplify.
WORKSPACE:
ANS:
a*a*a , a^2 , k*k*k , k^5 , a^4 , b^6 , x , y^7 , m-n , a*a*a*a*a*a*a , 4 , k*k*k*k*k*k , 2 , 1 a^3 , 1 b^5 , 1 x , 1 y^8 , n-m , x^-1 , x^-4 , x^-5 , x^-3 , m-n , x^-6 , 3^-4 , d^-2 , r^-7 , x^-8 , -n , n t^2 , z^5 y , 1 n^2k^2 , g^4h , n m^2 , 1 , 2 , 4 , 3 , 1 , 3 , 9 , 1 , 1 , r^3y , x^-5 , 1 , n^3m^-6 , x^-1 , y^-3 , p^2k^-4 , x^9 , 1
RATIONAL EXPRESSIONS: Problem Set 1
NOTES:
Problem 1:
A 1979 television commericial for Timex watches contained the statement that the watches were 99.97% accurate. In a day, how much time might be lost or gained by a watch with that accuracy? Round to the nearest second.
ANS: 26 seconds
***
In 1983 an economist stated that all of the world's population could easily fit in the state of Texas, with plenty of room for homes and yards. The world's population is 4.7 billion. The area of texas is 267,300 square miles. How many square feet would there be for each person?
3
ANS: 1.585 * 10 sq ft/person
(OR: 1,585 sq ft/person)
***
A 30-foot by 60-foot ballroom is to be turned into a discotheque by placing an 18-foot by 42-foot dance floor in the middle and carpeting the remainder. What percent of the floor is dance area? Round the nearest percent.
ANS: 42%
WORKSPACE:
ANS:
26 , 1.585 * 10^3 , 42
POLYNOMIALS AND EXPONENTIAL OPERATIONS II: Test
This lesson will evaluate your skills in dividing polynomials and simplifying exponential expressions.
NOTES:
WORKSPACE:
ANS:
*Editor's Note: Made some mistakes, taken out*
OPERATIONS ON RATIONAL EXPRESSIONS: Simplifying Rational Expressions
- Simplify rational expressions by canceling common factors
- Analyze graphs of rational functions
NOTES:
In this activity we will simplify rational expressions like:
2
x + 8x + 7
----------
2
x - 2x - 3
By finding common factors and canceling. We also will investigate how this affects the graph of the expression.
***
Let's talk about equivalence.
Sometimes things that are equivalent under one condition are not necessarily equivalent under another condition.
Let's look at an example...
Two halves are not always equivalent to one whole.
Suppose we are given these two halves of a coat.
*Picture of left and right side of coat torn in half*
We can say that the sum of the halves is equivalent to one whole coat.
However, suppose we are given these two halves.
*Picture of two left sides ^_^ *
We cannot say that the sum of these halves is equivalent to one whole coat.
So, under some conditions we can say that two halves are equivalent to one whole, and other conditions they are not.
***
A similar case arises in simplifying rational expressions.
2x 2(x) 2
------ = -------- = -----
2 x(x - 2) x - 2
x - 2x
According to the example above,
2x
------ should be equivalent to,
2
x - 2x
2
-----
x - 2
We will see that for some values of x the expressions are equivalent, but this is not true for ALL values of x.
*Shows picture of a graph impossible to reproduce*
These graphs are equivalent EXCEPT at x = 0. For the graph on the left there is a hole at x = 0, and for the one on the right there is not.
If we want the graph on the right to be equivalent to the one on the left we RESTRICT the expression.
2 /
----- x = 0
x - 2
By excluding the x-value where the hole occurs, our graph on the left and our graph on the right become equivalent.
***
Let's look at some more examples:
x + 1 x + 1 1
----- = -------------- = -----
2 (x + 1)(x - 1) x - 1
x - 1
Here are the graphs of these expressions.
*Sigh, can't show pictures...*
The hole in the left graph indicates that in order for these two expressions to be equivalent:
/
x = -1
***
Let's simplify this expression:
2
x + 5x + 6 (x + 2)(x + 3) x + 3
----------- = -------------- = -----
2 (x + 2)(x - 5) x - 5
x - 3x - 10
How must we restrict
x + 3
-----
x - 5
so that it is equivalent too
2
x + 5x + 6 /
----------- ? x = -2
2
x - 3x - 10
***
Before pressing ENTER, descirbe in your own words one way to make the simplified form of an expression equivalent to the original expression.
Look at the graphs of both expressions. If there are "holes" in the original graph, restrict the simplified expression at the x-value of the holes.
***
We can find the "holes" of a graph WITHOUT graphing!
*Yay!*
x - 3 x - 3
----------- = --------------
2 (x - 3)(x + 7)
x + 4x - 21
This expression has a hole in its graph at x = 3.
2
x - 2x - 8 (x + 2)(x - 4)
---------- = --------------
2 (x + 1)(x - 4)
x - 3x - 4
This expression has a hole in its graph at x = 4.
2
x - 4 (x - 2)(x + 2)
------ = --------------
3 x(x - 2)(x + 2)
x - 4x
This expression has holes in its graph at x = 2 and x = -2.
***
At what x-values are there "holes" in the graphs of the following expressions?
x + 2
---------
2
x + x - 2
There is a hole at x = ... -2.
*Next time factor it, sigh.*
***
Before pressing ENTER, describe in your own words one way to determine the values for the holes in the graph of an expression.
The holes occur at the x-values which make both the numerator and the denominator zero. This occurs because there are common factors in the numerator and the denominator.
***
In addition to restricting a simplified expression at the x-values where there are holes, we must also restrict it at those x-values that make it undefined.
x + 4 x + 4 1
---------- = -------------- = -----
2 (x + 4)(x - 3) x - 3
x + x - 12
/
x = -4
Notice: If x = 3, the expression is undefined since the denominator would be zero.
2x - 8 2(x - 4) 2
------ = -------------- = -----
2 (x + 4)(x - 4) x + 4
x - 16
/
x = 4, -4
***
Before pressing ENTER, describe in your own words the restricted values of a simplified rational expression.
The restricted values of the simplified expression are the values of x which make the denominator of the original expression zero.
WORKSPACE:
ANS:
-1 , x + 3 x - 5 , -2 , -2 , 1 , -3 , -2 , 1 , -3 , 4 , -5 , 3 x - 2 , -2 2 , x + 3 , 5 , x - 1 x + 2 , -6 -2 , 2x - 2 x + 8 , 3 -8 , x + 2 x - 1 , 4 1 , x + 5 x + 2 , -2 5 , 1 2x^3 , -3 -1 , x + 3 x + 1 , -3 -1 , x , 0 3 , x - 1 x + 9 , -1 -9 , x + 2 x - 2 , 2 4 , x + 4 x - 4 , 4 -5 , x + 1 x - 5 , 5 6 , x + 7 x^3 , 0 -2 , x - 8 x - 7 , 7 -7
The above is continued (some parts are from this date, though). Current Date: March 15, 2006.
OPERATIONS ON RATIONAL EXPRESSIONS: Multiplication and Division
- Multiply and divide rational expressions
NOTES:
In this activity we will use what we've learned about simplifying rational expressions to help us multiply and divide rational expressions.
Remember how we multiply rational numbers.
(We multiply numerator by numerator and denominator by denominator.)
3 8 24 2
- * - = -- = -
4 9 36 3
We could simplfy this process by canceling common factors first.
3 8 2 * 3 * 4 2
- * - = --------- = -
4 9 3 * 3 * 4 3
We follow the same process for rational expressions.
x + 4 x - 3 (x + 4)(x - 3) x + 4
----- * ----- = -------------- = -----
x - 3 x + 6 (x - 3)(x + 6) x + 6
Here we multiply and cancel common factors.
- We factor first on this one.
2 2
x + 4x - 5 x + 5x - 14
---------- * -----------
2 x + 5
x + 6x - 7
Equals...
(x + 5)(x - 1) (x + 7)(x - 2)
-------------- * --------------
(x - 1)(x + 7) (x + 5)
- Then cancel common factors.
- Finally we simplify.
= x - 2
***
Before pressing ENTER, describe in your own words one way to multiply rational expressions.
Factor the numerator and denominator of each expression and "cancel" like factors before multiplying.
***
Recall how we divide rational numbers.
3 9 3 4 12 1
- / - = - * - = -- = -
8 4 8 9 72 6
/\ \\
|| \\
We first Then we simplify.
invert and
multiply.
We could simplify this process by canceling common factors first.
/ /
3 9 3 4 3 * 4 1
- / - = - * - = ------------- = -
8 4 8 9 2 * 4 * 3 * 3 6
/ / /
We follow the same process for the rational expressions.
x + 3 2x + 6 x + 3 3x + 12
----- / ------- = ----- * -------
x + 4 3x + 12 x + 4 2x + 6
\\
\\
We invert
and multiply.
/////// ///////
(x + 3) * 3(x + 4) 3
= ------------------ = - We cancel
(x + 4) * 2(x + 3) 2 to simplify.
/////// ///////
We follow the same process for the rational expressions.
2
x - 2 x + 3x - 10
---------- / -----------
2 2
x + 2x - 3 x + 8x + 15
We invert
and multiply.
\\ 2
x - 2 x + 8x + 15
= ---------- * -----------
2 2
x + 2x - 3 x + 3x - 10
/////////////////////
(x - 2)(x + 5)(x + 3) 1
= ---------------------------- = -----
(x + 3)(x - 1)(x + 5)(x - 2) x - 1
/////// //////////////
- We cancel to simplify.
***
Before pressing ENTER, describe in your own words one way to divide rational expressions.
Change the quotient to a product by inverting the divisor, factor each expression completely, cancel common factors and simplify.
***
WORKSPACE:
3(x + 1) -2
-------- * -------
(x + 4) (x + 5)
*Note to self: Simplify ALL THE WAY*
***
ANS:
6 , -6x - 6 x^2 + 9x + 20 , 5 x - 4 , x + 4 x + 1 , 3x^2 - 9x - 30 2x - 2 , 1 3 , x + 2 x + 3 , x - 2 x + 4 , 3x - 15 2x - 8 , x + 3 x - 1 , x - 5 x + 4 , 9x - 18 2x - 6 , 1 1 , 1 1 , x^2 - 6x + 9 x^2 + 7x + 10 , x^2 - 7x + 6 x^2 - 4 , -x + 2 x - 1 , x^2 - 9x + 20 -2x^2 - 16x - 30 , x^2 - 2x + 1 x^2 + 4x + 3 , x + 1 x + 3 ,
The above is continued (some parts are from this date, though). Current Date: March 16, 2006.
OPERATIONS ON RATIONAL EXPRESSIONS: Addition and Subtraction
- Add and subtract rational expressions
NOTES:
In this activity we will extend what we learned about multiplying and dividing rational expressions to help us add and subtract rational expressions.
***
Remember how we add and subtract common denominators.
3 5 8
- + - = - = 4
5 2 2
We simply add the numerators and simplify.
14 2 12 4
-- - - = -- = -
9 9 9 3
Here we subtract the numerators.
***
We follow the same process for rational expressions with common denominators.
Addition:
2x + 9 4x - 7 (2x + 9) + (4x - 7)
------ + ------ = -------------------
2x - 4 2x - 4 2x - 4
6x + 2 3x + 1
= ------ = ------
2x - 4 x - 2
Subtraction:
5x + 3 4x - 5 (5x + 3) - (4x - 5)
------ - ------ = -------------------
3x + 1 3x + 1 3x + 1
x + 8
= ------
3x + 1
***
Before pressing ENTER, describe in your own words how we add and subtract rational expressions with common denominators.
We add or subtract the numerators and leave the denominator the same. We then simplify if possible.
***
Remember how we add and subtract rational numbers with different denominators.
5 2 5 7 2 4 35 8 43
- + - = - * - + - * - = -- + -- = --
4 7 4 7 7 4 28 28 28
- We find a common denominator.
- We add the numerators.
Another:
4 5 4 5
- - - = ----- - -----
9 6 3 * 3 2 * 3
- First we find a common denominator.
4 2 5 3
= - * - - - * -
9 2 6 3
- Next we subtract the numerators.
8 15 -7
-- - -- = --
18 18 18
We follow the same process for rational expressions with different denominators.
x - 6 x + 5
----- + -----
x + 4 x - 4
- Find the common denominator.
x - 6 (x - 4) x + 5 (x + 4)
----- * ------- + ----- * -------
x + 4 (x - 4) x - 4 (x + 4)
2 2
x - 10x + 24 x + 9x + 20
= -------------- + --------------
(x + 4)(x - 4) (x - 4)(x + 4)
- Add the numerators and simplify.
2
2x - x + 44
= -----------
2
x - 16
Another:
2
x + 3x + 2 1
---------- - -----
2 x + 3
x + 4x + 3
- We find a common denominator.
2
x + 3x + 2 1 (x + 1)
-------------- - ----- * -------
(x + 3)(x + 3) x + 3 (x + 1)
2
x + 3x + 2 - (x + 1)
--------------------
(x + 3)(x + 1)
- Then we subtract and simplify.
(x + 1)(x + 1) x + 1
-------------- = -----
(x + 3)(x + 3) x + 3
***
Before pressing ENTER, describe in your own words how we add and subtract rational expressions with different denominators.
We factor each denominator to determine the factors of the least common denominator. Then we change each expression so it has the same denominator and then simplify.
***
Sometime's it is best to reduce each expression BEFORE finding a common denominator.
2
x - 9 8x
----- + ------
x + 3 2
x + 2x
(x + 3)(x - 3) 8x
= -------------- + --------
(x + 3) x(x + 2)
x - 3 8
----- + -----
1 x + 2
(x - 3)(x + 2) 8
= -------------- + -----
(x + 2) x + 2
2 2
(x - x - 6) + 8 x - x + 2
= --------------- = ---------
x + 2 x + 2
Another:
2
x - 3x - 10 x + 6
----------- - -----------
x - 5 2
x + 3x - 18
(x + 2)(x - 5) (x + 6)
= -------------- - --------------
(x - 5) (x + 6)(x - 3)
1 (x + 2)(x - 3) 1
= x + 2 - ----- = -------------- - -----
x - 3 (x - 3) x - 3
2 2
(x - x - 6) - 1 x - x - 7
= --------------- = ---------
x - 3 x - 3
WORKSPACE:
x x - 1
----- - -----
x + 2 x + 3
x(x + 3) (x - 1)(x + 2)
-------------- - --------------
(x + 2)(x + 3) (x + 3)(x + 2)
2 2
x + 3x x + x - 2
---------- - ----------
2 2
x + 5x + 6 x + 5x + 6
***
4x 2
----- + -----------
x + 2 2
x + 7x + 10
4x 2
----- + --------------
x + 2 (x + 2)(x + 5)
4x 2
----- + --------------
x + 2 (x + 2)(x + 5)
2
4x + 20x + 2
------------
2
x + 7x + 10
***
2
3x - 4 x + 3x + 2
------ - ----------
x - 1 2
x - 1
2
3x - 4 x + 3x + 2
------ - ----------
x - 1 (x + 1)(x - 1)
2x - 6
------
x - 1
***
2
2x - 7x - 4 x + 3
----------- + -----
x - 4 x - 7
2
2x - 7x - 4 x + 3
----------- + -----
x - 4 (x - 7)(x - 4)
(2x + 1)(x - 4) (x + 3)
--------------- + -------
(x - 4) (x - 7)
(2x + 1)(x - 4) (x + 3)
--------------- + -------
(x - 4) (x - 7)
2
2x - 12x - 4
------------
x - 7
***
2 2
3x + x - 2 x + 5x - 24
---------- - -----------
2 x - 3
x + 9x + 8
(3x - 2)(x + 1) (x + 8)(x - 3)
--------------- - --------------
(x + 1)(x + 8) (x - 3)
2
-x - 13x - 66
-------------
x + 8
***
ANS:
-x + 1 2x + 1 , x + 4 , 12x - 11 x + 4 , 6x + 2 2x - 3 , 10x - 5 3x - 1 , 2x + 2 x^2 + 5x + 6 , 2x^2 - 6x + 3 x^2 + 2x - 15 , -2x - 1 x^2 - 5x - 6 , 4x^2 + 20x + 2 x^2 + 7x + 10 , 2x - 6 x - 1 , 2x^2 - 12x - 4 x - 7 , -x^2 - 13x - 66 x + 8 , x^2 + 16x + 48 x^2 + 3x - 54 , 9x + 54 x^2 - 8x - 20 ,
The above is continued (some parts are from this date, though). Current Date: March 20, 2006.