## Topics

1. How to determine that it is a quadratic equation
2. How to expand
3. How to factor
4. How to solve
5. How to complete the square

## What is a quadratic equation?

A quadratic equation is a polynomial equation that is equal to 0 or y

y =ax²+bx+c

This is standard form above.

In a quadratic equation the only value that can not be 0 is the 'a' value in the equation. This is because if the 'a' value is missing the equation is no longer quadratic, it is linear. Even if the 'b' and 'c' value are missing from the equation the equation will still remain quadratic.

Examples:

## How to Factor

In our lessons of factoring we learned how to factor using a method which is known as factoring by inspection.

This is usually the easiest way of finding your x-intercepts however it works mainly with simple questions.

When factoring you have a couple rules you must follow in order to get the correct answer

• You must always multiply the 'a' value and the 'c' value (even if there is just a "-x")

Factoring Simple trinomials

y = x²+5x+6 --------> Decomposition

y = x²+2x+3x+6

y =x(x+2) +3(x+2)

y =(x+2) (x+3)

Step 1:Muliply (1)(6)

Step 2:What two numbers are the product of 6 and the sum of 5?

Step 3:When you find that out you will have your two binomials immediately

How to Factor complex trinomials

y = -2x²+7x-5

y= -2x²+2x+5x-5

y= -2x(x-1) +5(x-1)

y= (x-1) (-2x+5)

Step 1:Multiply the 'a' value and 'c' value. In this case you will multiply (-2)(-5)= +10

Step 2: What two numbers are the product of 10 and the sum of 7?

Step 3: place those 2 numbers, (+5, +2) in the equation in the place of the 7x. This is because the two numbers have a total sum of your 'b' value.

Step 4: Collect like terms in pairs now, since you have 4 values of numbers

Step 5: you will end up with your two binomials

## How to Expand

Now that we know how to factor, lets check out if we factored correctly by expanding

In order to expand you need to multiply the brackets together

(x+3)(x+2)

x²+2x+3x+6

x²+5x+6

Step 1: Multiply both brackets

Step 2: Collect like terms

Now lets expand using another example

2(x+3)(x+2)

2(x²+2x+3x+6)

2(x²+5x+6)

2x²+10x+12

Step 1: Multiply your brackets because of the BEDMAS rule

Step 2: Collect like terms within the brackets

Step 3: Multiply the bracket with the number outside the bracket

You are now done

This is a test question and may confuse people so I will explain this

(x+3)²

(x+3)(x+3)

x²+3x+3x+9

x²+6x+9

Step 1: Square the bracket; By this I mean make another bracket with the same binomials it contains, hence in this case (x+3)² becomes (x+3)(x+3)

Step 2:Multiply the brackets together

You are now done

Solving by factoring

y=x²+5x+6

y=x²+2x+3x+6

y=(x+2) (x+3)

x=-2; x=-3

Step 1: First we factor you can learn how to factor using my lesson above

Step 2: Using your two binomials in brackets you will solve separately since in a quadratic equation there are two x-intercepts.

Step 3: Now you solve within the brackets

Step 4: y=x+2 | y=x+3

Step 5: Put your x on one side so that you find what x is equal to when y is 0

Examples:

1. x-1=0

x=1

2. x+9=0

x=-9

Quadratic formula is best to be used when you can not solve for your roots by factoring

-b±√b²-4ac

---------------

2a

Step 1: You have to know your 3 values (A,B,C)

Step 2: Plug your numbers into the equation

Step 3: Solve

*Key while solving: Make sure you divide by '2a' last. Divide once everything has been calculated* --- This is a must or your answer will be incorrect

Examples:

1. -2x²+8x+4

A: -2 B: 8 C: 4

1.

-(-2)±√8²-4(-2)(4)

-----------------------

2(-2)

2.

2±√64+32

--------------

-4

3.

2±√96

--------

-4

4.

2±9.8

--------

-4

x= -2.95 ; x= 1.95

## How to Complete the Square

From standard form, completing the square can also be a way to help you solve your quadratic. However completing the square is most convenient when trying to figure out the vertex form of the parabola. This helps you solve maximum and minimum word problems where it helps you find the peak in the graph.

y= -2x²+8x-4

y= -2(x²-4x)-4

y= -2(x²-4x+4-4)-4

y= -2(x²-4x)+8-4

y= -2(x-2)²+4

Step 1:Factor out the 'a' value and block out the 'c' value

Step 2:Add the zero, You divide by 2, and then square it -4/2 = -2² = 4

Step 3:Take the negative out of the equation and eliminate the positive # from the zero

Step 4:Solve what is outside on the right side of the bracket by either subtraction or addition. Also you divide your 'b' value by two and remove the x from the 'bx' value. After doing that, you remove the square from the 'x²' value and place it outside of the bracket