Download Quadratic Formula

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This tutorial is in two parts:
•Part One walks you through the procedures for
solving a quadratic equation using the Quadratic
Formula.
•Part Two describes the process of predicting the Type
of Solution you will get without actually having to use
the whole formula.
There is a chart at the end of the lesson which
is a guide for Predicting Types of Solutions.
The Quadratic
Formula
I. When to use the Quadratic Formula:
When factoring any equation that is in Quadratic Form
Quadratic Form:
2
ax
+ bx + c = 0
where a, b and c are numbers
II. How to use the Quadratic Formula:
Replace each a, b and c in the formula below with it’s value,
then simplify it as far as possible.
x=
b
b  4ac
2a
2
Example 1:
The coefficient
of the x2 term
is the “a”
2x2 - 5x + 7 = 0
The coefficient
of the x term is
the “b”
The constant
is the “c”
Here’s a tip to prevent sign errors:
First write the Quadratic Formula with
empty parentheses in place of each a, b and c.
2
-( )  ( ) - 4( )( ) c
x=
2( ) a
b
Notice: The “x”, the “4” and the “2” never change. Also, notice
where the two negatives and the “plus/minus” sign are located.
Lastly, notice that the first term inside the square root sign is
squared (the 2 is outside the parentheses).
Example 1:
2x2 - 5x + 7 = 0
Fill in the empty parentheses with the correct numbers
(including the sign) and simplify the equation
x=
2
(

5
)
 4(2)(7)
-(-5) 
b
2(2)
c
a
The first term becomes a +5 and the bottom is a 4
x=
5
4
We can simplify the square root piece on the side,
this will save us a lot of writing.
x=
(5)2  4(2)(7)
5
4
This section of the formula (the square root piece)
is called the discriminant.
25  (8)(7)
25  56

 31
 i 31
Now we can put this back into the fraction
Look at the numerator (the top of the fraction),
The 5 is a real number and the discriminant is imaginary.
This type of number is called a Complex Number.
This means the numerator can’t be simplified any further.
x=
5  i 31
4
The equation is now in simplified form.
(i.e. this is the answer)
But, finding the value of x is not the only question
you might be asked to answer.
You may be asked to describe the type of solution
you will get for x.
The solution to this equation is:
x=
5  i 31
4
Since the fraction has a “” symbol in it,
it is really two numbers.
x=
5 + i 31
4
and x =
5 - i 31
4
Type of Solution:
Two Complex Solutions were found for the quadratic equation
2x2 - 5x + 7 = 0
Example 2:
4x2 + 7x - 2 = 0
Write the equation with empty parentheses for a, b and c
2
-( )  ( ) - 4( )( ) c
x=
2( ) a
b
Fill in the empty parentheses with the correct numbers
(including the signs) and simplify the equation
-(7)  (7)  4(4)(2)
x=
2(4) a
b
2
c
The first term becomes a -7 and the bottom is an 8
-7  72  4(4)(2)
x=
8
Now simplify the discriminant
7 2  4( 4)( 2)

49  (16)( 2)

49  32
 81
9
Now we can put this back into the fraction
The solution to this equation is:
x=
-7  9
8
Since the fraction has a “” symbol in it, this is 2 numbers.
x=
-7 + 9
8
and
x=
-7 - 9
8
The numerator has two real numbers which can be combined.
2 1
 16
2
and x 
x 
   2
8 4
8
1
Type of Solution:
Two Real Number Solutions were found for the quadratic equation
4x2 + 7x - 2 = 0
Example 3:
3x2 - x - 6 = 0
Notice the “b” is a -1
Write the equation with empty parentheses for a, b and c
2
-( )  ( ) - 4( )( ) c
x=
2( ) a
b
Fill in the empty parentheses with the correct numbers
(including the signs) and simplify the equation
-(-1)  (1)2  4(3)(6)
x=
2(3) a
b
c
The first term becomes a 1 and the bottom is a 6
1  (1)  4(3)(6)
6
2
x=
Now simplify the discriminant
(1) 2  4(3)( 6)
 1  (12)( 6)
 1  72
 73
Now we can put this back into the fraction
The solution to this equation is:
x=
1  73
6
Since the fraction has a “” symbol in it, this is 2 numbers.
x=
1+
6
73 and
x=
1-
73
6
The numerator has two real numbers which cannot be combined.
Type of Solution:
Two Real Number Solutions were found for the quadratic equation
3x2 - x - 6 = 0
Example 4:
4x2 + 6x + 1 = 0
Write the equation with empty parentheses for a, b and c
2
-( )  ( ) - 4( )( ) c
x=
2( ) a
b
Fill in the empty parentheses with the correct numbers
(including the signs) and simplify the equation
x=
-(6) 
b
(6)  4(4)(1) c
2
2(4)
a
The first term becomes a 1 and the bottom is a 6
x=
-6 
(6)  4(4)(1)
2
8
Now simplify the discriminant
(6) 2  4(4)(1)
 36  (16)
 20
2 5
Now we can put this back into the fraction
The solution to this equation is:
x=
-6  2 5
8
Since the fraction has a “” symbol in it, this is 2 numbers.
-6 + 2 5 and
x=
8
x=
-6 - 2 5
8
The numerator has two real numbers which cannot be combined.
But, they can be REDUCED, each of these terms can be divided by two.
x=
-3 + 5
4
and
x=
-3 - 5
4
Type of Solution:
Two Real Number Solutions were found for the quadratic equation
4x2 + 6x + 1 = 0
Example 5:
4x2 + 4x + 1 = 0
Write the equation with empty parentheses for a, b and c
2
-( )  ( ) - 4( )( ) c
x=
2( ) a
b
Fill in the empty parentheses with the correct numbers
(including the signs) and simplify the equation
x=
-(4) 
b
(4)  4(4)(1) c
2
2(4)
a
The first term becomes a 1 and the bottom is an 8
x=
-4 
(4)  4(4)(1)
2
8
Now simplify the discriminant
(4)  4(4)(1)
2
 16  (16)
 0
0
Now we can put this back into the fraction
The solution to this equation is:
x=
-4  0
8
Since the fraction has a “” symbol in it, this is 2 numbers.
40
4
x

8
8
and
40
4
x

8
8
This is one real number repeated twice.
So, once we reduce it, the solution is:
4
1
x 
8
2
Type of Solution:
One Real Number Solution was found for the quadratic equation
4x2 + 4x + 1 = 0
Let’s do a recap:
We have seen many examples of how to use the quadratic formula.
The procedure was the same in each case:
1. We wrote the quadratic formula using
empty parentheses in place of the a, b and c.
2. We filled in the values of a, b and c, including the signs.
3. We simplified the equation.
4. We split the simplified equation into two pieces
to determine if it could be simplified further.
Next, we described the type of solution(s) we obtained.
We got one of the following three types of solutions:
1. Two real number solutions.
2. Two complex number solutions.
3. One real number solution.
Predicting the
Type of
Solution
The type of solution is depends on the value of the discriminant.
We can predict the type of solution by looking at the discriminant
before or after we simplify or reduce it.
Go back and look at all the examples we worked.
Example 1 had the solution:
5
 31
4
5  i 31
4
Discriminant is negative (i.e. imaginary).
Predict the Type of Solution: Two Complex Solutions
Actual
Solutions:
5  i 31
5  i 31
x
,x 
4
4
Example 2 had the solution:
 7  81
8
79
8
The numerator can be simplified further
Discriminant is positive (i.e. Real).
Predict the Type of Solution: Two Real Number Solutions
1
Actual
x  , x  2
Solutions:
4
Example 3 had the solution:
1
73
It won’t reduce
6
Discriminant is positive (i.e. Real).
Predict the Type of Solution: Two Real Number Solutions
Actual
Solutions:
1  73
1  73
x
,x 
6
6
Example 4 had the solution:
 6  20
8
62 5
8
The fraction can be simplified further
Discriminant is positive (i.e. Real).
Predict the Type of Solution: Two Real Number Solutions
Actual
Solutions:
3
x
2
5
3 5
,x 
2
Example 5 had the solution:
4
8
0
40
8
This becomes a single fraction
Discriminant is zero (i.e. Real).
Predict the Type of Solution: One Real Number Solution
Actual
Solution:
1
x
2
The chart below outlines how to predict the type of
solution(s) by looking at the discriminant alone.
Discriminant
Produces
Type of Solution
positive
real numbers
Two Real
Number Solutions
negative
imaginary
numbers
Two Complex
Number Solutions
zero
One Real Number
one real number
Solution
Practice Exercises
Part I: Predict the types of solutions you will get for the following equations.
[Two Real, Two Complex or One Real Number Solution(s)]
Equations
Predictions
2 Reals
1. 3x2- 2x - 1
2 Reals
2. 4x2+ 2x - 5
2 Complex
3. 2x2+ 3x + 4
1 Real
4. 9x2 - 6x + 1
2 Real
5. 3x2 + 5x + 2
2 Complex
6. 5x2- 2x + 1
Part II: Use the Quadratic Formula to find the actual solutions for
the six equations above. [See if your predictions were correct]
Solutions: 1. 1, -1/3
2.
 1  21
4
3.
 3  i 23
4
4. 1/3
5. -2/3, -1
6.
2  4i 1  2i

(completely reduced)
10
5
End of Tutorial
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