Math 3
3-1 Quadratics Revisited

Name ____________________________________
I can use my knowledge of quadratics from previous courses to solve problems
In Math 2, you learned about different ways to solve quadratic equations. You also studied the mathematical effects
of gravity and initial velocity as they pertain to Pumpkin Dropping and Punkin’ Chunkin’. One of the properties we
learned was that gravity exerts force on any free-falling object so that d, the distance fallen, will be related to time t
by the function d  16t 2 .
PUMPKIN DROPPING (PART 1)
Each year, physics students at Old Dominion University in Virginia drop pumpkins off the top of a building. If the
building is 80 feet tall, the height of the pumpkin can be modeled by the function d (t )  16t 2  80 . To find the
amount of time the pumpkin would be in free-fall, we could graph the function d(t).
1a.) Graph the function d(t).
1b.) Use your calculator to determine how long the pumpkin was airborne.
1c.) What is the height of the pumpkin the instant it hits the ground?
1d.) Another way of determining the amount of time the pumpkin is
airborne is to solve the function algebraically. Solve the equation
0  16t 2  80 .
When we set a quadratic function equal to zero and solve it, we are finding the x-intercepts.
2a.) Graph the function f ( x )  x 2  9 .
2b.) Solve 0  x 2  9 algebraically.
THE QUADRATIC FORMULA
Another way to solve quadratic equations is by using the Quadratic Formula. The Quadratic Formula is:
For equations of the form ax 2  bx  c  0 , x 
b
b2  4ac

2a
2a
Solve each equation using the Quadratic Formula. Graph it on your calculator to check your solution.
3a.) x 2  2 x  8  0
3b.) x 2  5 x  3  0
3c.) 3x 2  16 x  18  0
3d.) 2 x 2  7 x  9  0
3e.) What was unique about the graph of (3d)? What was unique about the solutions to (3d)?
IMAGINARY NUMBERS
In Math 2, you learned about a new set of numbers: the imaginary numbers. The square root of a negative yields an
imaginary solution. The basis for this was
1  i . Therefore,
16  4i , because
16  16  1  4i .
Simplify the following.
25  ______
9  ______
64  ______
28  ______
576  ______
4a.) Solve the following equation using the Quadratic Formula: 0  x 2  6 x  13
4b.) What do the solutions to the equation tell you about the graph of the equation?
FACTORING (PART 1)
In problem 3a, we solved using the Quadratic formula, to discover the x-intercepts were (2, 0) and (-4, 0). A faster
way to solve, when the x-intercepts are rational numbers, is by factoring. The key to solving using factoring is to set
the factors equal to zero. This works because of The Zero Product Property, which states that if the product of any
two quantities is zero, then one of the two quantities must equal zero  If A  B  0, then either A  0 or B  0
x2  2 x  8  0
( x  4)( x  2)  0
x40
x20
x  4
x2
Solve by factoring.
5a.) x 2  10 x  24  0
5b.) x 2  2 x  15  0
5c.) x 2  7 x  18  0
5d.) x 2  9  0
5e.) x 2  36  0
5f.) x 2  1  0
PUMPKIN DROPPING (PART 2)
Let’s say that the students at Old Dominion dropped a different pumpkin out of a window that was 64 feet above
the ground. The function would now become d (t )  16t 2  64 . If we attempt to solve for how long it would take
the pumpkin to hit the ground, we would solve the equation 0  16t 2  64 .
What if we factor out -16 from both terms in the equation? Can we solve it faster than the quadratic formula or the
factoring we used in 5d through 5e.?
6a.) Factor out 16 and solve using the Zero Product Property. 0  16t 2  64
6b.) Was one of your solutions negative? Does a negative solution make sense in the context of the problem?
Use the Zero Product Property to solve the following. Make sure to start by finding a common factor. Then graph
the equations on your calculator to check your solutions. Sketch the graphs below.
7a.) 2 x 2  32  0
7b.) x 2  3 x  0
7c.) 2 x 2  4 x  0
FACTORING (PART 2)
Factoring does not always lead to integer solutions (x-intercepts). In Math 2, you learned how to solve equations of
the form ax 2  bx  c  0 , when a  1 . In other words, you factored and solved equations like 6 x 2  17 x  5  0 ,
that had no common factors.
6 x 2  17 x  5  0
2x  5  0
3x  1  0
x  2.5
x
1
3
Solve the below problems by factoring. Graph the equations on your calculator to check your solutions.
2 x 2  13x  15  0
3x 2  10 x  8  0
Special Cases of Factoring
I.
Difference of Two Squares
Fill in the below table. Try to find a pattern.
Difference of Two (perfect) Squares
Factored Form
A.
x2  9
 x  3 x  3
B.
x 2  25
 x  5 x  5
C.
9 x 2  49
3x  7 3x  7 
D.
___________
 x  8 x  8
8x2  6 x  5  0
x 2  100
E.
_______________
 4 x  5 4 x  5
F.
G.
16 x 2  49
_______________
General Rule:
a 2  b2
_______________
II.
Perfect Square Trinomials
Fill in the below table. Try to find a pattern.
Perfect Square Trinomial
Factored Form
A.
x 2  8 x  16
 x  4
B.
x 2  8 x  16
___________
C.
___________________
 x  3
D.
x 2  10 x  25
___________
E.
x 2  18 x  _____
___________
F.
x 2  10 x  _____
___________
G.
x 2  5x  _____
___________
General Rule:
a 2  2ab  b 2
_______________
2
2
The trinomials above were examples of perfect square trinomials. Why do you think they are called that?
Now we can solve quadratic equations by applying these skills.
COMPLETING THE SQUARE.
A quadratic equation in standard form is written y  ax 2  bx  c
A quadratic in vertex form is written y   x  h   k , where  h, k  is the vertex of the parabola.
2
To change an equation from standard form to vertex form, we complete the square
Write y  2 x 2  8x  5 in vertex form and find the vertex
Method 1 – finding the vertex by completing the square
1.
Factor the leading coefficient (a) out of the first two terms
y  2  x2  4x   5
2.
Next, take half of the middle term (b) and square it, then add it within the parenthesis and subtract it
from outside the parenthesis – remember to distribute the a!




The b term is 4
Half of 4 is 2
2 squared is 4
Add 4 inside the parenthesis, subtract 2  4 from outside the parenthesis
y  2  x2  4x  4  5  8
2
3.
1 

Factor the parenthesis – it will always factor into  x  b  – and simplify the rest of the equation.
2 

y  2  x  2   3  vertex of the parabola is  2, 3
2
Note that completing the square will show us how a parabola is transformed from the parent function y  x 2 .
The original equation y  2 x 2  8x  5 can be written as y  2  x  2   3 , meaning that compared to y  x 2
2
the function y  2 x 2  8x  5 is translated to the left 2, stretched vertically by a factor of 2, and moved down 3.
You would never know that unless you completed the square!
OVER 
Method 2 – solving quadratic equations by completing the square
Solve 0  2 x 2  8 x  5 by completing the square
1.
Move the c term to the other side of the equation
5  2 x 2  8 x
2.
Factor out the leading coefficient (a) and divide through by it.

3.
5
 x2  4x
2
Next, take half of the middle term (b) and square it, then add it to both sides




The b term is 4
Half of 4 is 2
2 squared is 4
Add 4 to both sides
5
  4  x2  4 x  4
2
2
4.
1 

Factor the right side of the equation – it will always factor into  x  b  – and then isolate y
2 

3
3
3
3
2
2
  x  2 
  x  2  
 x  2  x  2 
2
2
2
2
The exact solutions are x  2 
3
and the approximate solutions are x  0.775 and x  -3.225
2
Practice: Solve by completing the square.
1.
x 2  18 x  31
LESSON 3-1 SUMMARY:
2.
x 2  10 x  19  0
3.
x 2  7 x  10