Download Solving Quadratic Equations: Three Methods

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Cubic function wikipedia , lookup

Fundamental theorem of algebra wikipedia , lookup

Factorization wikipedia , lookup

Quartic function wikipedia , lookup

Elementary algebra wikipedia , lookup

History of algebra wikipedia , lookup

System of linear equations wikipedia , lookup

Equation wikipedia , lookup

Quadratic equation wikipedia , lookup

System of polynomial equations wikipedia , lookup

Transcript
Name: __________________________
Solving Quadratic Equations: Three Methods
Method 1: Factoring
1) Review: Solve x2 + 6x + 8 = 0 by factoring in a generic rectangle.
Factors of 8:
Sum of the
Factors
x2
8
The solution set is {
}
2) Review: Solve 3x2 + 14x + 8 = 0 by factoring in a generic rectangle.
Factors of 3x8:
Sum of the
Factors
3x2
8
The solution set is {
Method 2: Completing the Square
3) Solve x2 + 8x + 10 = 0
You could always try to solve the quadratic by factoring. So you would need two numbers that multiply together to 10
and add the 8. Can you find any?
In our new method “completing the square” we will use a square to complete x2 + 8x into a perfect square. All though
we won’t use the algeblocks for this activity you can visualize x2 + 8x as follows:
x2
x x x x x x x x
This isn’t a square (yet). Let’s move half of the x-pieces underneath:
x2
x
x
x
x
x x x x
What is missing to
complete the square?
Remember that the short
side of the x-tile is 1.
The missing piece is: _____
}
Our completed square now has an equation of x2 + 8x + 16
Of course when we complete the square by adding an extra term we have to compensate for this and
subtract the same term to keep the original equation intact!
The equation was x2 + 8x + 10 = 0. If we add 16 we have to subtract 16 to keep the original equation balanced:
x2 + 8x + 16 + 10 - 16 = 0
When we factor the square above we get (x+4) (x+4) or (x+4)2. Let’s re-write our equation:
(x+4)2 +10 -16 = 0
Notice that we took half of 8
Remember:
(x+4)2 -6 = 0
which is 4 and then squared this
If 32 = 9 and (-3)2=9
Now you can solve this for x:
to complete our square with 16!
Then √9
3
The solution set is {
}
If an equation doesn’t factor nicely you can try solving it by completing the square! If there is a coefficient in
front of x2 divide out the coefficient before completing the square. We can use generic squares instead of
algeblocks to complete the square as well.
4) 4x2 -8x -1 = 0 Solve by completing the square. (Divide out 4 first)
Show your work:
x2
- 2x
x2
The solution set is {
}
Method 3: Quadratic Formula
In standard form ax2 + bx + c=0 can be solved by completing the square (see your e-textbook section 9.2 page
505 for a worked out version – aka proof ):
√
5) Solve 4x2 – 8x + 1 = 0 using the quadratic formula:
Label a =
b=
c=
√
x=
The solution set is {
}
The Discriminant
The discriminant of ax2 + bx + c = 0 is b2 – 4ac. We can calculate the discriminant to predict the number and
type of solutions.
Discriminant
If b2 – 4ac > 0
If b2 – 4ac = 0
If b2 – 4ac < 0
Number of (real #) Solutions
2 real # solutions
1 real # solution
0 real # solutions
Type of Solutions
Rational or Irrational
Rational
2 Complex Solutions
x-intercepts
2
1
0
exercises
Solve the following quadratics with one of the 3 methods. Show ALL work for full credit and use “completing the
square” in at least one problem. Use the discriminant to determine the number and type of solutions.
6) 3x2 – 72 = 0
Number of (real #) solutions:
Type of solutions:
The solution set is {
}
7) x2 + 5x -1 = 0
Number of (real #) solutions:
Type of solutions:
The solution set is {
}
8) 2x2 – 5 x – 2 = 0
Number of (real #) solutions:
Type of solutions:
The solution set is {
}
9) 6x2 – 5x -4 = 0
Number of (real #) solutions:
Type of solutions:
The solution set is {
}
10) 9x2 – 6x + 5 = 0
Number of (real #) solutions:
Type of solutions:
The solution set is {
}