Transcript Quadratic Equations

MTH 065
Elementary Algebra II
1
Chapter 11
0011 0010 1010 1101 0001 0100 1011
Quadratic Functions and Equations
2
4
Section 11.1
Quadratic Equations
Geometric Representation of
Completing the Square
0011 0010 1010 1101 0001 0100 1011
x
1
x+8
Area = x(x + 8)
2
4
Geometric Representation of
Completing the Square
0011 0010 1010 1101 0001 0100 1011
x
Area = x2 + 8x
2
x
8x
x
8
1
2
4
Geometric Representation of
Completing the Square
0011 0010 1010 1101 0001 0100 1011
x
Area = x2 + 8x
2
x
8x
x
8
1
2
4
Geometric Representation of
Completing the Square
0011 0010 1010 1101 0001 0100 1011
x
Area = x2 + 8x
2
x
8x
x
4 4
1
2
4
Geometric Representation of
Completing the Square
0011 0010 1010 1101 0001 0100 1011
x
Area = x2 + 8x
2
x
4x 4x
x
4 4
1
2
4
Geometric Representation of
Completing the Square
0011 0010 1010 1101 0001 0100 1011
x
Area = x2 + 8x
2
x
4x 4x
x
4 4
1
2
4
Geometric Representation of
Completing the Square
0011 0010 1010 1101 0001 0100 1011
4
4x
x
2
x
4x
x
4
Area = x2 + 8x
1
2
4
Geometric Representation of
Completing the Square
0011 0010 1010 1101 0001 0100 1011
4
4x
?
x
2
x
4x
x
4
Area = x2 + 8x + ?
1
2
4
Geometric Representation of
Completing the Square
0011 0010 1010 1101 0001 0100 1011
4
4x
16
x
2
x
4x
x
4
Area = x2 + 8x + 16
1
2
4
Geometric Representation of
Completing the Square
0011 0010 1010 1101 0001 0100 1011
4
4x
16
x
2
x
4x
x
4
Area = x2 + 8x + 16 = (x + 4)2
1
2
4
Terminology
• Quadratic Equation
0011 0010 1010 1101 0001 0100 1011
Any equation equivalent to an equation with the form …
ax2 + bx + c = 0
… where a, b, & c are constants and a ≠ 0.
• Quadratic Function
Any function equivalent to the form …
f(x) = ax2 + bx + c
... where a, b, & c are constants and a ≠ 0.
1
2
4
Review Results from Chapter 6
0011 0010 1010 1101 0001 0100 1011
• Solve quadratic equations by graphing.
• Put into standard form: ax2 + bx + c = 0
• Graph the function:
f(x) = ax2 + bx + c
• Solutions are the x-intercepts.
• # of Solutions? 0, 1, or 2
1
2
4
Details of Graphs of Quadratic Functions – Section 11.6
Review Results from Chapter 6
0011 0010 1010 1101 0001 0100 1011
• Solve quadratic equations by factoring.
• Put into standard form: ax2 + bx + c = 0
• Factor the quadratic: (rx + m)(sx + n) = 0
• Set each factor equal to zero and solve.
• # of Solutions?
2
• 0  does not factor (not factorable  no solution)
• 1  factors as a perfect square (if it factors)
• 2  two different factors (if it factors)
1
4
Principle of Square Roots
0011 0010 1010 1101 0001 0100 1011
For any number k, if …
x k
2
… then …
x 
k,  k
1
2
4
Principle of Square Roots
0011 0010 1010 1101 0001 0100 1011
For any number k, if …
x k
2
… then …
x k
Why? Consider the following example …
x2 = 9



x2 – 9 = 0
(x – 3)(x + 3) = 0
x = 3, –3
1
2
4
Application of the
Principle of Square Roots
0011 0010 1010 1101 0001 0100 1011
Solve the equation …
5 x  15  0
2
5 x  15
2
x 3
2
x 3
Note
This example
demonstrates how
to solve a quadratic
equation with no
linear (bx) term.
1
2
4
Application of the
Principle of Square Roots
0011 0010 1010 1101 0001 0100 1011
Solve the equation …
5 x  15  0
2
5 x  15
2
x  3
2
x   3  i 3
Note
Remember to always
simplify radicals.
1
2
• no perfect squares
• no multiples of perfect
squares
• no negatives
4
Application of the
Principle of Square Roots
0011 0010 1010 1101 0001 0100 1011
Solve the equations …
( x  3)  4
( x  5)  7
x  3  2
x5   7
x  3 2
x  5  7
2
x  5, 1
2
1
2
4
Application of the
Principle of Square Roots
0011 0010 1010 1101 0001 0100 1011
Solve the equation …
x  8 x  16  11
2
x  8x  5  0
2
But this does not factor …
( x  4)  11
2
x4  
x  4  11
1
11
2
4
Solving by
“Completing the Square”
0011 0010 1010 1101 0001 0100 1011
x  6x  7  0
2
x  6 x  7
2
2
x  6 x  3  7  9
Note: This polynomial
does not factor.
2
( x  3)  2
2
x3  2
x  3  2
1
2
4
2
ax
Solving
+ bx + c = 0 by
“Completing the Square”
0011 0010 1010 1101 0001 0100 1011
• Basic Steps …
1. Get into the form: ax2 + bx = d
2. Divide through by a giving: x2 + mx = n
3. Add the square of half of m to both sides.
2
m
• i.e. add  
2
4. Factor the left side (a perfect square).
1
2
4
5. Solve using the Principle of Square Roots.