Transcript 11.1 Solving Quadratic Equations by the Square Root Property • Square Root Property of Equations: If k is a positive number and if.

11.1 Solving Quadratic Equations
by the Square Root Property
• Square Root Property of Equations:
If k is a positive number and if a2 = k, then
a k
or
a- k
and the solution set is: {
k ,- k}
11.1 Solving Quadratic Equations
by the Square Root Property
• Example:
5 x  2
2
3
5 x  2  3 or 5 x  2   3
5 x  2  3 or 5 x  2  3
2 3
2 3
x
or x 
5
5
11.2 Solving Quadratic Equations
by Completing the Square
• Example of completing the square:
x  6 x  7  0 cannotbe factored
2
x  6 x  9  2  0 (completethesquare)
2
( x  3)  2  0  ( x  3)  2
2
2
x  3   2  x  3  2
11.2 Solving Quadratic Equations
by Completing the Square
•
Completing the Square (ax2 + bx + c = 0):
1. Divide by a on both sides
(lead coefficient = 1)
2. Put variables on one side, constants on the
other.
3. Complete the square (take ½ of x coefficient
and square it – add this number to both sides)
4. Solve by applying the square root property
11.2 Solving Quadratic Equations
by Completing the Square
2
2
x

y
 ( x  y )(x  y )
• Review:
x2  y2
(prime)
x 3  y 3  ( x  y )(x 2  xy  y 2 )
x  y  ( x  y )(x  xy  y )
3
3
2
2
x  y  ( x  y )(x  y )  ( x  y)(x  y)(x  y )
4
4
2
2
2
2
2
• x4 + y4 – can be factored by completing the square
2
11.2 Solving Quadratic Equations
by Completing the Square
• Example:
Complete
the square:
   y 
 x   2 x y  y   2 x y
 x  y    2 xy
x y  x
4
4
2 2
2 2
2

2 2
2
2

 x  y  2xy x  y  2xy
2
2
2
2 2
2
Factor the difference
of two squares: 2
2 2
2
2

11.3 Solving Quadratic Equations
by the Quadratic Formula
• Solving ax2 + bx + c = 0:
Dividing by a: x2  b x  c  0
a
a
Subtract c/a:
x  ba x   ac
2
Completing the
2
square by
x 
adding b2/4a2:
b
a
x
b2
4a 2
 
c
a
b2
4a 2
11.3 Solving Quadratic Equations
by the Quadratic Formula
• Solving ax2 + bx + c = 0 (continued):
2
Write as a
2
b
 4ac
b 2
4 ac
b
square: x  2 a    4 a 2  4 a 2  4a 2
Use square root
property:
b  b2  4ac
x

2a
2a
 b  b2  4ac
Quadratic formula: x 
2a
11.3 Solving Quadratic Equations
by the Quadratic Formula
• Quadratic Formula:
 b  b  4ac
x
2a
2
b  4ac
2
is called the discriminant.
If the discriminant is positive, the solutions are
real
If the discriminant is negative, the solutions are
imaginary
11.3 Solving Quadratic Equations
by the Quadratic Formula
• Example:
x2  5x  6  0
x
a  1, b  -5, c  6
 (5) 
5
( 5)  4(1)(6)
2(1)
2
25  24 5 1

 
2
2 2
x  3, x  2
11.3 Solving Quadratic Equations
by the Quadratic Formula
•
Complex Numbers and the Quadratic Formula
Solve x2 – 2x + 2 = 0
 (2)  (2)  4(1)(2)
x
2(1)
2
2   4 2  4i 2  2i



2
2
2
 1 i
11.4 Equations Quadratic in Form
Method
Advantages
Factoring
Fastest method
Disadvantages
Not always
factorable
2
Square root
Not always this
form : ( x  a)  b
property
form
Completing the Can always be Requires a lot
square
used
of steps
Quadratic
Can always be Slower than
Formula
used
factoring
11.4 Equations Quadratic in Form
•
Sometimes a radical equation leads to a quadratic
equation after squaring both sides
•
An equation is said to be in “quadratic form” if it
can be written as a[f(x)]2 + b[f(x)] + c = 0
Solve it by letting u = f(x); solve for u; then use
your answers for u to solve for x
11.4 Equations Quadratic in Form
•
Example:
   4x
x  4x  3  x
4
2
2 2
2
3 0
Let u = x2
x 
 4 x  3  0  u   4u  3  0
(u  3)(u  1)  0  u  3, u  1
2 2
2
2
x 2  3, x 2  1  x   3 , x  1
11.5 Formulas and Applications
•
Example (solving for a variable involving a
square root)
4A
Solve : d 
for A

d 
2
4A

square bot h sides
d 2  4 A mult iplybot h sides by 
d 2
4
 A divide bot h sides by 4
11.5 Formulas and Applications
•
Example:
s  2t 2  kt
solvefor t
0  2t 2  kt  s
get zero on right side
-k  k 2  4( 2)( s )
t
2(2)
(quadrat ic formula)
-k  k 2  8s

4
-k  k 2  8s
-k  k 2  8s
so t 
and t 
4
4
11.6 Graphs of Quadratic Functions
•
•
A quadratic function is a function that can be
written in the form:
f(x) = ax2 + bx + c
The graph of a quadratic function is a parabola.
The vertex is the lowest point (or highest point if
the parabola is inverted
11.6 Graphs of Quadratic Functions
•
Vertical Shifts:
f ( x)  x  k
2
The parabola is shifted upward by k units or
downward if k < 0. The vertex is (0, k)
•
Horizontal shifts:
f ( x)  x  h
2
The parabola is shifted h units to the right if h >
0 or to the left if h < 0. The vertex is at (h, 0)
11.6 Graphs of Quadratic Functions
•
Horizontal and Vertical shifts:
f ( x)  x  h  k
2
The parabola is shifted upward by k units or
downward if k < 0. The parabola is shifted h
units to the right if h > 0 or to the left if h < 0
The vertex is (h, k)
11.6 Graphs of Quadratic Functions
•
Graphing:
f ( x)  ax  h  k
2
1. The vertex is (h, k).
2. If a > 0, the parabola opens upward.
If a < 0, the parabola opens downward (flipped).
3. The graph is wider (flattened) if 0  a  1
The graph is narrower (stretched) if a  1
11.6 Graphs of Quadratic Functions
Inverted Parabola with Vertex (h, k)
f ( x)  x  h  k
2
Vertex = (h, k)
11.7 More About Parabolas;
Applications
•
Vertex Formula:
The graph of f(x) = ax2 + bx + c has vertex
 b

,
 2a
  b 
f
 
 2a  
11.7 More About Parabolas;
Applications
•
Graphing a Quadratic Function:
1. Find the y-intercept (evaluate f(0))
2. Find the x-intercepts (by solving f(x) = 0)
3. Find the vertex (by using the formula or by
completing the square)
4. Complete the graph (plot additional points as
needed)
11.7 More About Parabolas;
Applications
•
Graph of a horizontal (sideways) parabola:
The graph of x = ay2 + by + c or x = a(y - k)2 + h
is a parabola with vertex at (h, k) and the
horizontal line y = k as axis. The graph opens to
the right if a > 0 or to the left if a < 0.
11.7 More About Parabolas
Horizontal Parabola with Vertex (h, k)
x  y  k   h
2
Vertex = (h, k)