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Ch8 Quadratic Equation Solving
Methods
General Form of Quadratic Equation
ax2 + bx + c = 0
A quadratic Equation: x2 – 7x + 10 = 0
1
-7 c = ______
10
a = _____
b = _____
Methods & Tools for Solving Quadratic Equations
1. Factor
2. Apply zero product principle (If AB = 0 then A = 0 or B = 0)
3. Square root method
Example1:
Example 2:
4. Completing the Square x2 – 7x + 10 = 0
4x2 – 2x = 0
5. Quadratic Formula
(x – 5) (x – 2) = 0
2x (2x –1) = 0
x – 5 = 0 or x – 2 = 0
2x=0 or 2x-1=0
+5 +5
+2 +2
2 2
+1 +1
2x=1
x = 5
or
x= 2
x = 0 or x=1/2
8.1 Square Root Method
Solving a Quadratic Equation with the Square Root Method
Example 1:
Example 2:
4x2 = 20
(x – 2)2 = 6
4
4
x – 2 = +6
x2 = 5
+2 +2
x = + 5
So, x = 5 or - 5
x = 2 + 6
So, x = 2 + 6 or 2 - 6
You try one. What is different about this one? : 2x2 + 1 = 0
Completing the Square
If x2 + bx is a binomial then by adding b 2 which is the square of half
2
the coefficient of x, a perfect square trinomial results:
x2 + bx + b 2 = x + b 2
2
2
Solving a quadratic equation with ‘completing the square’ method.
Example:
x2 - 6x + 2 = 0
-2 -2
x2 - 6x = -2
x2 - 6x + 9 = -2 + 9
(x – 3)2 = 7
x – 3 = + 7
x = (3 + 7 ) or (3 - 7 )
Step1: Isolate the Binomial
Step 2: Find ½ the coefficient of x (-3 )
and square it (9) & add to both sides.
Note: If the coefficient of x2 is not 1 you
must divide by the coefficient of x2 before
completing the square. ex: 3x2 – 2x –4 = 0
(Must divide by 3 before taking ½ coefficient of x)
Step 3: Apply square root method
8.2 Quadratic Formula
General Form of Quadratic Equation:
ax2 + bx + c = 0
Quadratic Formula: x = -b + b2 – 4ac
2a
discriminant: b2 – 4ac
if 0, one real solution
if >0, two unequal real solutions
if <0, imaginary solutions
Solving a quadratic equation with the ‘Quadratic Formula’
2x2 – 6x + 1= 0
2
a = ______
-6
b = ______
x = - (-6) + (-6)2 – 4(2)(1)
2(2)
= 6 + 36 –8
4
= 6 + 28
4
= 6 + 27
4
= 2 (3 + 7 ) = (3 + 7 )
4
2
1
c = _______
8.5 & 8.6 Quadratic Functions &
Graphs
y=x
2
x
0
1
-1
2
-2
y
0
1
1
4
4
A Parabola
Vertex
Lowest point if
The parabola opens upward,
And highest point if parabola
Opens downward.
Do you know what an axis of symmetry is?
Quadratic Functions & Graphs
y = x2 - 2
x
0
1
-1
2
-2
y
-2
-1
-1
2
2
Vertex
Notice this graph is shifted down 2 from the origin.
Y = x2 – k (shifts the graph down k units)
Y = x2 + k (shifts the graph up k units)
To shift the graph to the right or to the left
y = (x – h)2
(shifts the graph to the right)
y = (x + h)2
(shifts the graph to the left)
General Form of a Quadratic
y = ax2 + bx + c
a, b, c are real numbers & a 0
A quadratic Equation: y = x2 + 4x + 3
1
4
3
a = _____
b = _____
c = ______
x
-5
-4
-3
-2
-1
0
1
Where is the vertex?
Where is the axis of symmetry?
y
8
3
0
-1
0
3
8
Formula for
Vertex:
X = -b
2a
Plug x in to
Find y
The Role of “a”
a0
a0
f ( x)  x 2
f ( x)   x 2
f ( x )  2 x
2
f ( x)  2 x
2
f ( x )  3 x
2
f ( x)  3x
2
Quadratic Equation Forms
• Standard Form:
2
y  ax  bx  c
• Vertex Form:
y  a ( x  h)  k
2
Vertex = (h, k)
Examples
Find the vertex, axis of symmetry, and graph
each
a.
y  3( x  5)  2
b.
y  2( x  2)  3
Vertex (-2, -3)
1
2
y  ( x  4)  6
2
Vertex (4, -6)
c.
2
2
Vertex (5, 2)
Convert from Vertex Form to
Standard Form
Vertex Form:
y = 2(x + 2)2 + 1
To change to standard form, perform multiplication,
add, and combine like terms.
y = 2 (x + 2) (x + 2) + 1
y = 2 (x2 + 2x + 2x + 4) + 1
y = 2 (x2 + 4x + 4) + 1
y = 2x2 + 8x + 8 + 1
y = 2x2 + 8x + 9
(Standard Form)
Convert from Standard Form to
Vertex Form
(Completing the Square – Example 1)
Step 1: Check the coefficient of the x2 term. If 1 goto step 2
If not 1, factor out the coefficient from x2 and x terms.
Step 2: Calculate the value of : (b/2)2
Step 3: Group the x2 and x term together, then add (b/2)2 and subtract (b/2)2
Step 4: Factor & Simplify
Example 1: y = x2 –6x – 1 (Standard Form)
y = (x2 –6x + 9) – 1 -9
y = (x – 3) (x – 3) – 1 – 9
y = (x – 3)2 – 10 (Vertex Form)
(b/2)2 = (-6/2)2 = (-3)2 = 9
Convert from Standard Form to
Vertex Form
(Completing the Square – Example 2)
Step 1: Check the coefficient of the x2 term. If 1 goto step 2
If not 1, factor out the coefficient from the x2 and x term.
Step 2: Calculate the value of : (b/2)2
Step 3: Group the x2 and x term together, then add (b/2)2 and subtract (b/2)2
Step 4: Factor & Simplify
Example 2: y = 2x2 +4x – 1 (Standard Form)
y = 2( x2 + 2x) –1
(2/2)2 = (1)2 = 1
y = 2(x2 +2x + 1) – 1 -2 (WHY did we subtract 2 instead of 1?)
y = 2(x + 1) (x + 1) – 1 – 2
y = 2(x + 1)2 – 3 (Vertex Form)
Solving Quadratic Equations
General Form of a Quadratic Equation
y = ax2 + bx + c
0 = ax2 + bx + c (If y = 0, we can solve for the x-intercepts)
A quadratic Equation: y = x2 + 4x + 3
1
4
3
a = _____
b = _____
c = ______
Graphical Solution
Numerical
Solution
Vertex (-2, -1)
Formula for
Vertex:
X = -b
2a
x
-5
-4
-3
-2
-1
0
1
y
8
3
0
-1
0
3
8
Symbolic/Algebraic Solution
x2 + 4x + 3 = 0
(x + 3) (x + 1) = 0
x+3 =0
x+1=0
x = -3
x = -1
Number of Solutions
y = x2 + 4x + 3
2 Real Solutions
x = -3
x = -1
y = x2 - 4x + 4
1 Real Solution
x=2
y = 2x2 + 1
NO Real Solutions
(No x-intercepts)
An Application
3 in
Height of a picture = X
Length of picture = X + 5
The frame is 3 inches thick on all sides.
3
If the overall area of the picture and frame is 336 sq inches,
find the dimensions (height and length) of the picture.
3
x
x+5
3 in.
Arearectangle = Length x Height
Length = x + 5 + 6 = x + 11
Height = x + 6
336 = (x + 6) (x+11)
336 = x2 +11x +6x +66, so, x2 +17x – 270 = 0
(x – 10) (x + 27) = 0
x = 10 or x = -27
So, the picture is 10 inches by 15 inches
Another Application
Two cars left an intersection at the same time, one heading
Due north, the other due west. Some time later they were exactly 100
Miles apart. The car headed north had gone 20 miles farther than the car
Headed west. How far had each car traveled? (P. 269)
Leg12 + leg22 = Hypotenuse2
X2 + (x+20)2 = 1002
X2
+ X2
North
+ 40x + 400 = 10000
100
X + 20
2X2 + 40x + 400 = 10000
2X2 + 40x - 9600 = 0
X2 + 20x - 4800 = 0
(x + 80) (x – 60) = 0
X = -80 or x = 60
West
x
Intersection
Things to Know for the Quiz
1. Standard Form of a quadratic: y = ax2 + bx + c
Vertex Form of a quadratic: y = a(x – h)2 + k
• Find and graph the vertex
• Use an x/y chart to plot points & graph the parabola
• Show and give the equation for the line of symmetry.
• Convert from standard to vertex form and vice vera
2. Solving a quadratic equation to find its x-intercepts using these methods:
• Factor and apply zero product principle
• Square Root Method
• Completing the Square
Use when the equation won’t factor easily
• Quadratic Formula
3. Formulas I will give you:
• Quadratic Formula
• Formula for vertex of a standard form quadratic: x = -b
2a