Ch8 Quadratic Equation Solving Methods

General Form of Quadratic Equation

ax2 + bx + c = 0

A quadratic Equation: x2 – 7x + 10 = 0 a = _____ b = _____ c = ______

Methods & Tools for Solving Quadratic Equations• Factor • Apply zero product principle (If AB = 0 then A = 0 or B = 0)• Square root method• Completing the Square• Quadratic Formula

Example1: Example 2:x2 – 7x + 10 = 0 4x2 – 2x = 0(x – 5) (x – 2) = 0 2x (2x –1) = 0x – 5 = 0 or x – 2 = 0 2x=0 or 2x-1=0 + 5 + 5 + 2 + 2 2 2 +1 +1

2x=1x = 5 or x = 2 x = 0 or x=1/2

1 -7 10

8.1 Square Root Method

Solving a Quadratic Equation with the Square Root MethodExample 1: Example 2:4x2 = 20 (x – 2)2 = 64 4

x – 2 = +6 x2 = 5 + 2 + 2

x = + 5 x = 2 + 6

So, x = 5 or - 5 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: Step1: Isolate the Binomialx2 - 6x + 2 = 0 -2 -2 Step 2: Find ½ the coefficient of x (-3 )x2 - 6x = -2 and square it (9) & add to both sides.x2 - 6x + 9 = -2 + 9(x – 3)2 = 7x – 3 = + 7

x = (3 + 7 ) or (3 - 7 )

Note: If the coefficient of x2 is not 1 you must divide by the coefficient of x2 beforecompleting 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 FormulaGeneral Form of Quadratic Equation: ax2 + bx + c = 0

Quadratic Formula: x = -b + b2 – 4ac discriminant: b2 – 4ac 2a if 0, one real solution if >0, two unequal real solutions if <0, imaginary solutionsSolving a quadratic equation with the ‘Quadratic Formula’

2x2 – 6x + 1= 0 a = ______ b = ______ c = _______

x = - (-6) + (-6)2 – 4(2)(1) 2(2)

= 6 + 36 –8 4

= 6 + 28 = 6 + 27 = 2 (3 + 7 ) = (3 + 7 ) 4 4 4 2

2 -6 1

8.5 & 8.6 Quadratic Functions & Graphs

y = x2

x y0 01 1-1 12 4-2 4

A Parabola

Vertex

Lowest point ifThe parabola opens upward,And highest point if parabolaOpens downward.

Do you know what an axis of symmetry is?

Quadratic Functions & Graphs

y = x2 - 2

x y0 -21 -1-1 -12 2-2 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 a = _____ b = _____ c = ______1 4 3

x y-5 8-4 3-3 0-2 -1-1 0 0 3 1 8

Where is the vertex?Where is the axis of symmetry?

Formula forVertex:

X = -b 2a

Plug x in toFind y

The Role of “a”

2

2

2

3)(

2)(

)(

xxf

xxf

xxf

2

2

2

3)(

2)(

)(

xxf

xxf

xxf

0a0a

Quadratic Equation Forms• Standard Form:

• Vertex Form:

khxay 2)(

cbxaxy 2

Vertex = (h, k)

Examples

Find the vertex, axis of symmetry, and graph each

a.

b.

c.

2)5(3 2 xy

3)2(2 2 xy

6)4(2

1 2 xy

Vertex (5, 2)

Vertex (-2, -3)

Vertex (4, -6)

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)

Example 1: y = x2 –6x – 1 (Standard Form) (b/2)2 = (-6/2)2 = (-3)2 = 9

y = (x2 –6x + 9) – 1 -9

y = (x – 3) (x – 3) – 1 – 9

y = (x – 3)2 – 10 (Vertex Form)

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

Convert from Standard Form to Vertex Form

(Completing the Square – Example 2)

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)

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

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 a = _____ b = _____ c = ______1 4 3

x y-5 8-4 3-3 0-2 -1-1 0 0 3 1 8

Formula forVertex:

X = -b 2a

Vertex (-2, -1)

NumericalSolution

Graphical SolutionSymbolic/Algebraic Solution

x2 + 4x + 3 = 0

(x + 3) (x + 1) = 0

x + 3 = 0 x + 1 = 0

x = -3 x = -1

Number of Solutionsy = 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

Height of a picture = XLength of picture = X + 5The frame is 3 inches thick on all sides.

If the overall area of the picture and frame is 336 sq inches,find the dimensions (height and length) of the picture. 3 in.

3 in

3 3 x

x + 5

Arearectangle = Length x Height

Length = x + 5 + 6 = x + 11Height = 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 headingDue north, the other due west. Some time later they were exactly 100Miles apart. The car headed north had gone 20 miles farther than the carHeaded west. How far had each car traveled? (P. 269)

Intersection

North

West

100

x

X + 20

Leg12 + leg22 = Hypotenuse2

X2 + (x+20)2 = 1002

X2 + X2 + 40x + 400 = 10000

2X2 + 40x + 400 = 10000

2X2 + 40x - 9600 = 0X2 + 20x - 4800 = 0(x + 80) (x – 60) = 0X = -80 or x = 60

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

1. Find and graph the vertex2. Use an x/y chart to plot points & graph the parabola3. Show and give the equation for the line of symmetry.4. Convert from standard to vertex form and vice vera

2. Solving a quadratic equation to find its x-intercepts using these methods:1. Factor and apply zero product principle2. Square Root Method3. Completing the Square4. Quadratic Formula

3. Formulas I will give you:1. Quadratic Formula2. Formula for vertex of a standard form quadratic: x = -b 2a

Use when the equation won’t factor easily