# Quadratic Review- Algebraic Section

### By: Alina Farooq

## This review will go over Expanding, Factoring, Solving and Completing the Square

## What are Quadratic Equations?

For example: y=2x²+6x+8

This equation is in standard form

## Main Parts of the Parabola

## Expanding

Example

(x+a)(x+b)

=x²+bx+ax+ab

=x²+ a+b(x)+ab

First we have multiplied the terms.We multiplied the right side with x and then with a.

Now lets try an example with numbers

(x+2) (x+3)

**STEP 1**

**=x²+3x+2x+6** -We multiplied the first x to (x+3) which gave us x²+3x. Then we did the same with the 2 to (x+3) which gaves us 2x+6. That gave us the equation of x²+3x+2x+6

**STEP 2**

**=x²+5x+6-** We added like terms to simplify which gave us x²+5x+6

## Now lets try some examples

=x²+5x+6x+30

=x²+11x=30

b) (2x+3) (4x+20

= 8x²+4x+12x+6

= 8x²+16x+6

c) -2(3x+4)(2x-1)

= (-6x-8)(2x-1)

= -12x²+6x-16x+8

= -12x²-10x+8

## FACTORING

## There are many methods of factoring which will all lead us to solving for the variables. I will be showing you five methods. Common Factoring, Factoring by Grouping, Product Sum, Decomposition, and Difference of Squares

## Common Factoring

Example:

21x²+28xy

= 7x(3x+28y)

7x is the greatest common factor.

To check if your answer is correct you can expand your answer and you should get the same equation as it is in the beginning.

24+16x-8x²

=8(3+2x-x²)

x²+4x

=x(x+4)

## Product Sum Method

x²+5x-36

(p=-36) (s=5) You take the b value as the sum, and the c value as the product. Using trial and error you find two numbers that add and multiply to those numbers. The numbers are 6 and -1

=(x+6) (x-1)

## Decomposition

4x²-5x+6 We will multiply the a value and the c value

Product= 24 Sum=-5 Now we find two numbers that meet these value

The numbers are -8 and 3

4x²-8x+3x+6 We sub in the numbers and then factor by grouping

4x(x-2) 3(x-2)

(4x+3) (x-2) That leaves us with our final answer

## Difference of Squares

If you have an equation where there is squared value a minus and then another squared value that can be easily square rooted you can use this method.

Example

x²-81

=(x+9) (x-9)

49x²-25x²

(7x+5y) (7x-5y)

x^4-1

(x²+1) (x²-1)

(x-1)(x+1)(x²+1)

## Solving

## Solving is actually quite easy after you have factored an equation. We will learn how to solve with a factored equation and solve using the Quadratic Formula.

(x+2) (x-3)

We want to set this equation to zero so we put it into 0=(x+2)(x-3)

You then see what values would need to be plugged in for x for it to equal to zero

The numbers we are left with are x= -2 and x=3

Those values are the two x intercepts of the parabola

## Quadratic Formula

x = [ -b ± √(b2-4ac) ] / 2a

Lets plug x²-x-6=0 into the formula

a= 1 b= -1 c=6

After plugging those numbers into the formula the answers we are left with are x=3 x=-2

Those are the two x intercepts

Be careful if there is a negative under your square root that means that there are no x intercepts

## Completing the Square

This method is a multiple step process

1.(-5x²+20x)-2 Block off first 2 terms

2. -5(x²+4x)-2 Factor out the A

3. -5(x²+4x+4-4) -2 Take middle term divide by two and square

4. -5(x²+4x+4) 20-2 Take out negative term

5. -5(x+2)²+18 Factor

In this situation the x is the x value of the vertex and the 18 is the y value

## Word Problems

## Now lets try solving some word problems with our new knowledge

## Example 1

To solve this we must set the height to zero and solve for t(time)

0=-5t²+5t+30

= -5(t²-t+6)

Because this is not factorable we must plug it into the quadratic formula

That leaves us with the answers x=-2 x=3

The answer has to be positive so therefore the ball hits the ground in three seconds

## Example 2

To solve this question we must complete the square.

P=162x-81x²

=(-81x²+162)

= -81(x²-2+1-1)

=-81(x²-2+1) +81

=-81(x-1)²+81

We are going to use the y value

The maximum profit the company can make is 81 thousand dollars.

## Hopefully you now have a clearer understanding of quadratic equations.

Period 4

Mr. Ly