Quadratic Equations
And
Functions
By Chris Posey and Chris Bell

9-1 Square Roots
You already know how to find the square of a number i.e. 4²=16
To find the square root, you basically do the opposite of squaring a number. i.e. √(16)=4

Apply Skills Learned 
=12
• Now that you know how find the square root, let’s try some problems for practice!
• Find the square roots of the following problems.
1. √(144) 2. √(16) 3. √(400) 4 √(36)
1. √122 2. √16=4 =20 3. √400 4. √36=6

Example 1> x 2 =4
Write the original equation x= + √4 or - √4
Find square roots x= 2 or -2 2 2 =4 and (-2 2 )=4
Example 2> 3x 2 -48=0
3x 2 =48 x 2 =16 x= √16 x=4 or -4
Write the original equation
Add 48 to each side
Divide each side by 3
Find square roots
4 2 =16 and (-4) 2 =16

Apply Skills Learned 
• Now that you know how to solve quadratic equations by finding square roots, find the answers to the following problems.
1.5t
2 -125=0 2. x 2 =225 3. x 2 -15=10
1. t=5 2. x=15 3. x=5

• Product Property of Radicals: √ab= √a ∙ √b
Example> √50= √(25∙2)= 5 √2
Example> √48= √(4∙12)= 4 √3
• Quotient Property of Radicals: √a/b= √a/√b
Example> √32/50= √(2∙16)/√(2∙25)
Factor using square roots
=√(16/25)
=√(16)/√(25)
Divide common factors
Use Quotient Property
=4/5
Simplify

Apply Skills Learned 
Now that you know how to simplify radicals, try it out on your own!
1.
√(9/49)
2.
√(18)
3.
√(196)
4 √2 3)1 1)3/7 2)3

• A quadratic function is a function that can be written as a formula
Y=ax 2 +bx+c, where a ≠0
This will give a graph with a u shape called a parabola. If a is grater than 0, it opens up. If it is negative, then it opens down.

Graphing (9.4)
• Find the x coordinate of the vertex, which is x=-b/2a
• Make a x,y table and use the x values
• Plot the points and connect them with a smooth curve to form the parabola

Example
• Sketch graph of y=x 2 -2x-3
• Find x coordinate of vertex.
-b/2a=-2/2(1)=1
• Make a table x| -2 y|5
-1 0
0
1 2 3
-3 -4 -3 0
(1,-4) is the vertex, plot the rest of the points and draw a curve connecting them.
4
5

Practice
• Find vertex coordinates, and make a x,y value table using x values to the right and left of the vertex.
• y=-4x 2 -4x+8
9 8 0 -16
0 1 2 -.5
0 8
-1 -2 y|-16
9) x|-3
(-1/2,

• Solutions for the quadratic graphs are the x-axis intercepts, where y=0.
• This number can be checked in the original equation, by setting it equal to 0.

Example
•y=x 2 -2x-3 is shown here.
Note that the x intercepts are located at -1 and 3.
•If substituted for x, the equation would result in zero, the solutions for the equation.

Practice
• Solve the equation algebraically, check your answers by graphing.
• 2x 2 +8=16
±2 are tions Solu

• The solutions of the quadratic equation, ax 2 +by+c= 0, are
( -b+/√ ( b 2 4 ac ) x = ---------------------
(2 a ) when a ≠0 and b 2 4 ac≥ 0

Example
• Solve x 2 +9x=14=0
• Solution 1 x+ 9 x+ 14 =0
( -b+/√ ( b 2 4 ac ) x = ---------------------
(2 a )
( -( 9 )+/√ (( 9 ) 2 4( 1 )( 14 )) x = -------------------------------
(2( 1 ))
-9+/√(25) x=--------------------
2

-9+/-5
X=----------------
2
There are 2 solutions x=-2, and x=-7

Practice
• Solve the quadratic equation.
• 2x 2 -3x=8
9 =-1.3
d x , an X=2.89

• The discriminant is the radical expression in the quadratic formula ie. ( -b+/√ ( b 2 4 ac ) ) x = ---------------------
(2 a )
If the discriminant is positive, then the solution has
2 solutions.
If it is zero, it has one solution.
If it is negative, there are no real solutions.

Example
• Find value of the discriminat and determine if it has two solutions, one solution, or no solutions.
• x 2 -3x-4=0
• Use the equation, ax 2 +bx+c=0 to identify values, ie. a=1, b=-3, c=-4
• Substitute into the discriminant b 2 -4ac=(-3) 2 -4(1)(-4)
=9+16
=25
Discriminant is positive, therefore two solutions.

Practice
• Determine whether the graph will intersect the x-axis at one, two, or zero points.
• y=x 2 -2x+4 is the ax tersect t in es no It do

• The graph of a quadratic inequality consists of all the points (x,y) that are part of the inequality.
Quadratic inequalities can be represented by; y > or < or
 or
 ax 2 +bx+c

• When graphing, use a dashed line for the parabola when the equality is > or<
• Use a solid line when it is  or

• The parabola separates the graph into two sections. A test point is a point that is not on the graph.
If the test point is a solution, then shade the region, if not shade the other region.

Example
• Solve – x 2 + 4 < 0.
• Find the x-axis intercepts
– x 2 + 4 = 0 x 2 – 4 = 0
( x + 2)( x – 2) = 0 x = –2 or x = 2
• Use the origional inequality to find the area to shade
• y<0, therefore shade everything outside the parabola, below the x-axis.

Practice
• Determine whether the orderd pair is a solution of the inequality.
• y≤x 2 +7, (4,31) a rabol the pa tside of ou lution t is a so Poin