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 ,