Download C: Vertex Form

pg 1
4.1 Quadratic Functions and Transformations
A: Quadratic Functions
A QUADRATIC function is an equation in the form:
y  ax 2  bx  c
The _________ of a quadratic equation is a PARABOLA.
Ex. Circle the parabolas. Cross out the others.
y  2 x 2  15 x  18
y  6 x  9
y  x2  4x  4
y  16 x  22
B: Translations of x2
All parabolas are a translation of the parent graph y  x 2
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We discussed translations in chapter 2, lesson 2.6.
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Use your graphing calculator to graph the following translations and
state the transformations to the parent function.
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Ex. 2. y  3( x 1) 2  5
Ex. 1. y  2( x  3) 2 1
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Ex. 3. Match the graph with the equation
A
B
C
D
C: Vertex Form
Quadratic functions can also be written in “vertex form”:
y  a ( x  h) 2  k
The Vertex = ____________
Domain: the set of all x values for the equation
For an up/down parabola, usually:
The Axis of Symmetry = ____________
______________________________
“a” = _________________ of opening
Range:
positive: opens _______
the set of all y values for the equation
For an up/down parabola, usually:
negative: opens _______
_______________________________
Min/Max: The highest or lowest “y” value on
the graph, the y coordinate of the ___________
Ex. 4 Use your graphing calculator to graph:
y  2( x  2)2  5
State:
Vertex: __________
Min or Max? ______
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Value:_____
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A.O.S.: _________
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Domain: _________________
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Range: ___________________
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Ex. 5 Use your graphing calculator to graph:
y  3x 2  4
State:
Vertex: __________
Min or Max? ______
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Value:_____
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A.O.S.: _________
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Domain: _________________
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Range: ___________________
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WITHOUT A GRAPHING CALCULATOR:
Use a t-chart to graph the following. State the vertex and the equation of the axis of symmetry.
Ex. 6
y  ( x  5)2  2
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Vertex: _______
Make It Happen:
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A.O.S.: _______
1. Determine the
vertex. Plot the
point.
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2. Determine the A.O.S.
Draw in the vertical
line.
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3. Pick a few x-values
for your t-chart that
are to the right of
the A.OS.
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Ex. 7
y  2( x  4)2  6
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4. Use your t-chart to
find 2 points on the
graph.
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Vertex: _______
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A.O.S.: _______
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5. Use your A.O.S. to
reflect these points
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6. Sketch in the curve.
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D: Using Vertex Form  Find an Equation from a graph
Recall that vertex form: y  a( x  h)2  k
(h,k) is the vertex
(x,y) is any point on the curve
Example 8: Find the equation that models the graph at the right.
Step 1: What is the VERTEX: ______
so h = _____ and k = ______
Step 2: What is a POINT on the graph?
_______
so x = _____ and y = _______
Step 3: Substitute into vertex form and solve for a.
y  a ( x  h) 2  k
Step 4: Plug a, h, and k back into your vertex form.
Example 9: Find the equation that models the graph
Use these to FIND: ____
pg 5
4.2 Standard Form of a Quadratic Function
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A: Standard Form, Finding a Vertex
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If a quadratic function is ALREADY in vertex form y  a( x  h)2  k ,
you can find the vertex just by looking at the equation.
If a quadratic function is in STANDARD FORM y  ax 2  bx  c ,
you have to do a little work first to find the vertex.
Ex. 1 Use a graphing calculator to graph
y   x2  2 x  1
Vertex: ______
Min/Max? _______ Value: ______
Ex.
y   x2  2 x  1
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A.O.S.: ______
Domain: ________________________________
Finding a vertex
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Range: _________________________________
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Find the vertex, the axis of symmetry, the minimum or maximum value, the domain, and the range.
Ex. 2
y  2 x2  8x  1
Ex. 3
y  3x 2  6 x  9
Ex. 4 WITHOUT A GRAPHING CALCULATOR
Use a t-chart to graph y  x 2  2 x  3
State the vertex and the axis of symmetry
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B: Standard Form  Vertex Form
y  ax2  bx  c  y  a( x  h)2  k
Ex. 5 Rewrite y   x 2  4 x  5 in vertex form.
Step 1: Identify a & b
Step 2: Find the x coordinate of the vertex.
This value = h
Step 3: Substitute in the x-coordinate to find
the y-coordinate.
This value = k
Step 4: Write out vertex form.
Step 5: Substitute a, h, and k (Simplify if needed)
Rewrite in vertex form
Ex. 6
y  2 x 2  10 x  7
Ex. 8 Which is the graph of y  3x 2  4 x  6 ?
Ex. 7
y  x 2  16 x  66
pg 8
4-3 Modeling with Quadratic Functions
A: What is a Model? How do I make one?
A model is a way of using math to
describe a real world situation.
To model a quadratic equation, you
need to know the values of a, b, and c
in y  ax 2  bx  c .
Ex. 1
What is the equation of the parabola that contains the points (0,0), (1,-2), and (-1,-4)?
Make a Plan
 We need a, b, c
 We HAVE three points (x, y)
Make it Happen
 Substitute each of the three points (x,y) into y  ax 2  bx  c
 Solve the 3x3 systems of equations (like section 3.5)
 Substitute a, b, & c into y  ax 2  bx  c
y  ax 2  bx  c
Use (0,0)
Use (1,-2)
Use (-1,-4)
pg 9
X
0
2
1
Y
3
5
6
Ex. 2 Write the equation of the quadratic equation that passes through the points on the table.
B: Quadratic Regression
Sometimes your data points will not be EXACTLY a parabola, but the
basic shape of the data is parabolic. We can use a graphing calculator
to ESTIMATE the equation of a parabola that models the data.
Ex. 3 What is the quadratic model for
the data? (start by changing to a
24 hr clock!)
Quadratic Regression in Graphing
Calculator
1. Press the STAT key
2. Choose Option 1 “EDIT”
3. Enter your “x” data in the L1 list
Enter your “y” data in the L2 list
4. Press the STAT key
5. Arrow right to the CALC menu and
choose #5 Quad Reg
6. Enter
7. Use the estimated a, b, &c to write your
equation
pg 10
Ex. 4 Use quadratic regression to find a quadratic model for the
data in the chart. X represents the number of years since 1985.
a = _________________________
x=0
b = ________________________
x=5
x=15
c = ________________________
x=19
Equation: _______________________________________________
Use the equation to estimate the number of subscribers in 1995.
Ex. 5
Use quadratic regression to find a quadratic model for the data in the chart.
x represents the number of years since 1976
a = _____________________
x=0
b = _____________________
x=10
x=20
c = _____________________
Equation: ___________________________________________________
Use the equation to estimate the price per gallon in 2006 (x = 30).
x=29
pg 11
The main goal of Lessons 4.4-4.7 is to ____________ quadratic equations.
This means we are looking for values of x ( _____________________) that make the equation ax 2  bx  c  0
true.
y
For example:
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The solutions to x 2  x  6  0 are x  2 and x  3 .
Look at the graph: y  x 2  x  6
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What do you notice about x  2 and
x  3 ?
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Plug these values in to confirm.
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4.4 FACTORING
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What is factoring?
 A very useful algebra skill that we will use in NEARLY EVERY CHAPTER FOR THE REST OF THIS COURSE
 A way to “undo” the distributive property or the “FOIL” process
 A method of taking a quadratic expression like ax 2  bx  c and writing it as a PRODUCT of two
FACTORS
LOOK FOR _______________________ !
In this section we will REVIEW several factoring techniques:
1.
2.
3.
4.
Greatest Common Factor
x 2  bx  c
ax 2  bx  c when a  1
Difference of Two Perfect Squares a 2  b 2
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A: Greatest Common Factor
Recall Distributive Property
Multiply
The first step in any factoring problem is to look for the Greatest
Common Factor for all of your terms and “undistribute”.
2 x(3x  5)
Your final answer will look like the GCF (leftovers).
Examples: FACTOR!
A1.
7 x 2  21
A2.
11x 2  33 x
A3.
4 x3  x 2  16 x
A4.
3x 2  3x
A5.
9 x 3  45 x 2
A6.
12 x 2  15
B: Trinomials
ax 2  bx  c
a  1
After you’ve checked for a Greatest Common Factor.
Check to see if you have a trinomial with an “a”= 1 or “a” = -1.
You should try to find two special numbers that have a
Recall “FOIL” Process
Multiply
( x  2)( x  3)
PRODUCT of “c”
mxn=c
SUM of “b”
m+n=b
Factors:
(x + m) ( x + n)
Examples: FACTOR!
B1.
x 2  9 x  20
B2.
x 2  14 x  72
B3.
x 2  17 x  30
pg 13
B4.
 x 2  13x  12
C: Trinomials
ax 2  bx  c
a  1
Recall “FOIL” Process
Multiply
(2 x  3)( x  5)
B5.
 x 2  14 x  32
x 2  11x  30
B6.
After you’ve checked for a Greatest Common Factor.
After you’ve checked for a Trinomial with a = 1 or a = -1.
Look to see if you have a Trinomial with a = some other number.
You are going to have to use guess and check to “DE-FOIL”
ax 2  bx  c
(hx  m)( gx  n)
a  h g
b  hn  gm
c  m n
the FIRST Terms
the OUTSIDE + INSIDE terms
the LAST terms
Examples: FACTOR!
C1.
2 x 2  11x  12
C2.
4x2  4 x  3
C3.
2 x2  7 x  6
C4.
5x2  7 x  6
pg 14
D: Perfect Square Trinomials
Look for a pattern like this:
Multiply:
(2 x  3) 2
a 2  2ab  b2  (a  b)2
a 2 2ab  b2  (a  b)2
Essentially you can do these like part (C), but if you recognize
that the first term and the last term are perfect squares, it can
make the problem much easier to factor!
Examples: Factor!
D1.
4 x 2  24 x  36
E: Difference of Two Perfect Squares
Multiply:
( x  6)( x  6)
D2.
64 x 2  16 x  1
Look for a pattern like this:
a 2  b2  (a  b)(a  b)
Essentially you can do these like part (C), but if you recognize
that the first term and the last term are perfect squares, it can
make the problem much easier to factor!
Also be careful, unless there is a GCF
Examples: Factor!
E1.
x 2  81
E2.
x2 1
E3.
9x2  4
E4.
16 x 2  100 y 2
a 2  b 2 does not factor!
pg 15
F: Factor Completely
1. Check for GCF
2. Factor what’s leftover
3. Check each factored term to see if it will factor again!
Examples: Factor Completely
F1.
2 x 2  50
F2.
2 x 2  32 x  128
F3.
5 x 2  25 x  70
F4.
3x 2  9 x
F5.
2 x 2  22 x  60
F6.
12 x 2  10 x  12
F7.
7 x 2  700
F8.
18 x 2  12 x  2
F9.
 x 2  12 x  35
pg 16
4.5 Quadratic Equations
A: Solve by Graphing
Vocab:
A solution to a the quadratic equation ax 2  bx  c  0 is a value of x that makes the equation true.
A zero of a function f ( x)  ax 2  bx  c is a value of x that makes f(x) = 0.
y
An x-intercept of the graph of f ( x)  ax 2  bx  c is the x value on the graph where y  f ( x)  0 . This is the
place where the graph crosses the x-axis.
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For real numbers, “Solution” “Zero” and “X-intercepts” are basically synonyms and can be used interchangeably.
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For the equation y  x 2  x  6 whose graph is shown:
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The solutions to x 2  x  6  0
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The zeros of f ( x)  x 2  x  6
The x-intercepts of y  x 2  x  6
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Finding a “zero” in a graphing calculator
Given an equation ax 2  bx  c  0
1. Enter y  ax 2  bx  c
2. Press “Graph”
3. Look at your graph. How many zeros are there?
4. Press 2nd  Trace (Calc). Choose option 2 “Zero”
5. The calculator will ask you for the left bound.
Decide which zero you are looking for first.
Type in an x value to the left of your zero. Press enter.
6. The calculator will ask for a right bound.
Type in an x value to the right of your zero. Press enter.
7. The calculator will ask you to guess where you think the zero is.
Just press enter again.
8. The calculator tells you the zero of the function.
Repeat steps 4-8 for any additional zeros.
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Examples: Use your graphing calculator to find the zeros of the following functions. Give each answer to at
most two decimal places.
A1.
3x 2  24 x  45  0
A2.
3x 2  2 x  2  0
B: Solve by Factoring
You can also find the zeros of a quadratic function algebraically using factoring and the zero product property.
Make it Happen! Solve by Factoring!
Examples: Solve by Factoring
B1.
B3.
x2  5x  6  0
x 2  2 x  24  0
B2.
B4.
x 2  7 x  12
4 x2  6 x  0
1.
Make sure your equation
looks like ax 2  bx  c  0
2.
Factor completely.
3.
Set each factor = 0.
4.
Solve for x.
B5.
3x 2  45  24 x
pg 18
B6.
3x 2  x  4
B7.
25 x 2  20 x  4  0
B8.
9 x 2  49
B9.
3x 2  4 x  32
B10.
10 x 2  3 x  4
B11.
2 x 2  32  0
pg 19
4.6 Completing the Square
Quick Review! Simplifying Square Roots
A square root essentially asks the question: “What number times itself gives me this value?”
If your value is a perfect square, then that number will be an integer.
Examples:
49
144
If your value is NOT a perfect square, then you need to simplify the radical using the multiplication property.
a b  a  b
Examples:
Break your value down into factors,
one of which is a perfect square.
Also, don’t forget:
100
75
is NOT A REAL NUMBER.
A: Solve by Finding Square Roots
Examples: Solve
A1.
4 x 2  10  46
72
A2.
3 x 2  5  25
?
2
45
 100 is impossible for real numbers!
Make it Happen!
If your ax 2  bx  c  0 equation has no “bx” term,
you can solve by taking the square root of both sides of
the equation.
1. Use algebra to isolate the x 2
x2  n
2. Take the square root of both sides of the equation.
Don’t forget the  !
x n
3. Write as two equations.
x n
and x   n
pg 20
A3.
2 x 2  9  13
A4.
3 x 2  96  0
B: Perfect Square Trinomials
Recall that a trinomial in the form:
a 2  2ab  b2  (a  b)2
and
a 2  2ab  b2  (a  b)2
We can solve this form of equation using square roots as well!
Examples: Solve
B1.
( x  1)2  49
B2.
( x  8)2  20
B4.
4 x 2  20 x  25  17
Sometimes you will need to factor 1st!
B3.
x 2  14 x  49  25
pg 21
C: Completing the Square
Even if you don’t have a perfect square trinomial, you can still solve using this process using a technique called
“completing the square”
Here’s How it Works
Given a polynomial:
Example: Complete the square!
x 2  bx
C1.
2
Add
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 
2
Factor
b
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x 
2
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x 2  10 x
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x 2  14 x
2
x2  2 x
x2  5x
C3.
C4.
Using completing the square to solve quadratic equations
Make it happen:
Examples: Solve by Completing the Square
1. Isolate the
C5.
x 2  bx
x2  4x  2  0
b
2. Add  
2
2
to both sides
b
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3. Simplify (Factor)  x  
2
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4. Solve using square roots.
C6.
x 2  10 x  16  0
C7.
x 2  18 x  64  0
2
pg 22
D: Vertex Form
Completing the square can be useful for converting a standard form parabola into vertex form.
D1.
y  x 2  10 x  9
D2.
y  x 2  18x  13
pg 23
4.7 The Quadratic Formula
The quadratic formula is derived by taking the standard form of the quadratic equation: ax 2  bx  c  0 and
solving for x using the completing the square process.
Example: Solve using the quadratic formula
2 x2  5x  3  0
pg 24
The portion of the
equation under the
radical is called the
discriminant.
The discriminant tells
us a lot about the
types of answers we
might get.
Examples:
a) Find the discriminant.
b) Solve using quadratic formula
c) How many REAL solutions did you get?
Are they rational or irrational?
1.
3x 2  8 x  16  0
2.
4x2  x  2  0
3.
9 x 2  12 x  4  0
4.
x2  5x  8  0
pg 25
Remember that graphically, a real solution to a quadratic equation represents an x-intercept of the graph of the
equation.
Positive
Perfect Square:
Rational (ex. 3, ½, -4, etc)
Not Perfect Square:
Irrational (ex. 2  33 , etc.)
Zero
Negative
pg 26
Concept Byte: Writing Equations From Roots
Recall:
A SOLUTION is a value of x that makes the quadratic equation ax 2  bx  c  0 true.
Sometimes this is also called a ZERO of a function or a ROOT of an equation.
Synonyms
SOLUTION =
ZERO = ROOT
On a graph
A “REAL”
SOLUTION = ZERO = ROOT
2 x 2  11x  5  0
x  5 x  14  0
( x  7)( x  2)  0
2
Recall: “SOLVE BY FACTORING”
x7 0
x  7
X-Intercept
(2 x  1)( x  5)  0
or
x2 0
x2
2x 1  0
2 x  1
x
x5  0
x  5
1
2
We can work backwards from the roots (the solutions) to find the original equation.
Example: Write a quadratic equation with each pair of
values as roots.
Steps:
1. If “c” is a root, then (x –c ) is a factor.
2. Write your roots as the product of two
factors = 0.
1.
-7, 2
2.
-5, - ½
roots: b, c
equation: ( x – b) (x – c) = 0
3. Multiply using FOIL and simplify.
4. Multiply through by a constant to eliminate
fractions if needed.
pg 27
3.
2, 10
4.
5.
2
, -9
3
6.
-5, 5
3 1
 ,
4 2
Putting it all together 4.4-4.7
4.4 Factoring Patterns:
Greatest Common Factor
Trinomial
Trinomial a = 1
16 x 2  20 x
x 2  29 x  100
Trinomial a ≠ 1
3x 2  17 x  10
Difference of Perfect Squares
“Factor Completely”
x 2  64
 x 2  3x  28
25 x 2  4
Perfect Square
25 x 2  30 x  9
20 x 2  80
pg 28
4.5 Solve by Factoring
Solve by Factoring
Steps:
1. Get ax 2  bx  c  0
x 2  144  0
x 2  6 x  40  0
9 x 2  18 x  0
2x2  9 x  4  0
2. Factor
3.
Set each factor = 0
4. Solve
4.6 Solve by Completing the Square
Solve by Completing the Square
Steps:
1. Isolate x 2  bx
b
2
2
2. Add   to both sides
3. Factor and simplify
4. Solve using square roots
x2  4x  5  8
x 2  2 x  35  0
pg 29
4.7 Solve by Quadratic Formula
Solve by Quadratic Formula
Steps:
If
Then
ax 2  bx  c  0
x
2x2  x  4  2
b  b2  4ac
2a
8x2  1  7 x
Discriminant:
Steps: Find
b 2  4ac
Positive  2 real solutions
Zero 
1 real solutions
Negative  0 real solutions
Find the Discriminant. State the number of real solutions.
10 x 2  x  9  0
3 x 2  6a  3  0
9 x 2  6 x  11  0
pg 30
4.8 Complex Numbers
A. Imaginary Numbers
25
Consider:
Has no REAL solution because (?)2= -25
Imaginary Numbers were created to solve equations that did not have real solutions.
The imaginary unit “i” is used to represent:
i  1
and
i 2  1
Ex. Simplify the following
1.
36
2.
5.
90
6.
9.
2i  3i
10.
100
3.
7
3 8
7.
2  6
7i  3i  2i 
11.
 5i 
Numbers like 3i , 97i , and 7i are called PURE IMAGINARY NUMBERS.
2
4.
50
8. 3 10  4 3
12.
 3i 
3
pg 31
B: COMPLEX NUMBERS
A COMPLEX NUMBER has a real part “a” and an imaginary part “bi”.
A complex number looks like:
a  bi
Working with complex numbers:
Treat i like any other variable…
1.
(3  4i)  (7  5i)
2.
(9  8i)  (7  10i)
3.
(7  6i )  (6  10i)
4.
(17  20i)  (12  30i)
5.
(5  3i)(4  6i)
6.
(8  7i)(2  5i)
7.
7i (9  3i)
8.
4i (20  30i )
9.
(3  5i ) 2
10.
(4  2i)(4  2i)
11.
(6  5i)(6  5i)
Numbers in the form
a  bi and a  bi
are called complex conjugates.
pg 32
C: Solving Equations
Remember that a quadratic equation with no real solutions has a graph that has no x-intercepts.
This type of quadratic equation does have SOLUTIONS, though… COMPLEX SOLUTIONS.
SOLVE!
1.
x2  9  0
4.
3x 2  x  2  0
2.
5 x 2  20  0
3.
x 2  15  0
Quadratic Formula for
ax 2  bx  c  0
b  b2  4ac
x
2a
5.
x2  4x  5  0