# Quadratics

### Review Package

## Quadratic Relations

- Sometimes a curve of best fit is more appropriate model for data than a line of best fit.
- For constant increments of the independent variable (x), a relation is
**quadratic**if the second differences of the dependent variable (y) constant.*Example:*In a table of values the first differences are not constant but the second differences are constant. - A polynomial of degree 2 models a quadratic relation.
- A linear relation models a phenomenon where the rate of change is constant.
- A non-linear relation models a phenomenon where variable is the rate of change.

## Factoring

__COMMON FACTOR__

- Factoring is the opposite of expanding,
**ab+ac = a(b+c)** - If every term of a polynomial is divisible by the same constant, the constant is called a common factor.
- A polynomial is not considered to be completely factored until the Great Common Factor (GCF) has been factored out.

**Factor:** ab+ac

= a(b+c) , 4x+20

= (x+5)

- 4x+20

= 2(2x+10) (Not completely Factored) - 4x+20

= 4(x+5) (Completely Factored)

**EXAMPLES:**

- 3x-15y

= 3(x+5y) - 4a^3b^4 - 6a^2b^2 + 2ab

= 2ab(2a^2b^3 - 3ab+1)

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__FACTOR BY GROUPING__

- This is done by grouping a pair of terms. Then, factor each pair of two terms.
- x^2 + 2x^2 + 8x+16

= x^2(x+2) +8(x+2)

= (x+2) (x^2+8) - 8am-3bn-6an+4bm

= 4m(2a-b) -3n(b+2a)

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__SIMPLE TRINOMIALS__

- Simple trinomial equations are in the form of,
**x^2+bx+c** - In solving trinomials, the brackets have to be expanded (Multiply each term in the first bracket to each term in the second bracket).
- Equations in the form of
**x^2+bx+c**are called*Quadratic Equations*

**Expand: **x^2+bx+c ,** **(x+2) (x+1)

- (x+2) (x+1)

= x^2+3x+2 - "c" value is the product. Find out 2 numbers that add up to the "b" value and when multiplied the answer is the "c" value.

**EXAMPLES:**

- x^2 +7x +6 [Product=6 , Sum=7]

= (x+1) (x+6) - x^2 +7x -30 [Product=-30 , Sum=7]

= (x+10) (x-3) - x^2 +3x +5 [Product=5 , Sum=3]

=Not Factor-able

**Note: In some cases, first you might have to factor out. ***Example: **

Q: -5x^4 +5x^3 +30x^2 [First factor and then solve as a Simple Trinomial]

=-5x^2(x^2-x-6) [Product=-6 , Sum=-1]

=-5x^2(x-3)(x+2)

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__COMPLEX TRINOMIALS__

- Factor by Decomposition.
- Complex Trinomial are in the form of,
**Ax^2 +Bx +C.** - The "C" value is multiplied by the A value, only the coefficient.
- Determine the factors of the product which also add up to the "B" value.

**Factor by Decomposition: Ax^2 +Bx +C**

- 2x^2 +3x -5 [Product: A*C= -5*2=-10 , Sum=3]

=2x^2 +5x-2x -5 [Decompose the Middle term]

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

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

**EXAMPLES:**

- 14x^2 -19x -3 [Product: A*C= 14*-3=-42 , Sum=-19]

=5x^2 +10x+2x +4 [Decompose the Middle term]

=5x(x+2) +2(x+2)

=(5x+2) (x+2) - x^2 +7x +12 [Product: A*C= 12*1=12 , Sum=7]

=x^2 +3x+4x +12 [Decompose the Middle term]

=x(x+3) +4(x+3)

=(x+1) (x+3) - 5x^2 +10x+2x+4

=5x(x+2) +2(x+2)

=(5x+2) (x+2)

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**FACTORING DIFFERENCE OF SQUARES**

- The "a" value is squared (x^2)
- In the form of, (a+b) (a-b)
- Square-root of "b" value

**Expand:**

- (x+2) (x-2)

=x^2 - 4

- x^2 -16

= (x^2 - 4)

**EXAMPLES:**

- (a-3) (a+3)

=a^2 +3a-3a -9

=a^2 - 9 - (p-5) (p+5)

=p^2 - 25 - 4x^2 - 22

=2(2x^2 - 11)

## COMPLETING THE SQUARES

- Used to change Standard form equations to Vertex form

**STEPS TO COMPLETE THE SQUARES:**

- Block off the first two terms
- Factor out the A value only
- Divide the middle term by 2 and square it (b/2)^2
- Add and subtract the squared value answer
- Take out the negative squared value, Multiply the "a" value by the negative squared value when taking it out of the bracket
- Write within brackets x + the square root of the number and square the entire bracket and solve what is outside the brackets

**EXAMPLES:**

- x^2 + 6x - 2

=(x^2 + 6x) - 2

=(x^2 + 6x + 9 - 9) -2

=(x^2 + 6x + 9) -9 - 2

=(x+3)^2 - 11 - 3x^2 + 24x - 11

=3(x^2 + 8x) - 11

=3(x^2 + 8x + 16 -16) - 11

=3(x^2 + 8x +16) -48 -11

=3(x^2+4)^2 - 59

## QUADRATIC FORMULA

- In the form of, ax^2 + bx + c = 0
- Formula is used to identify the "x" value
- Quadratic formula is used when the equation cannot be factored
- Quadratic formula give 2 values of "x"
- Quadratic Formula:
x = -b ± √b^2 - 4ac /2a

**EXAMPLES:**

- x^2 - x + 8 = 0

x= -b ± √b^2 - 4ac /2a

x= 4x^2 + 33x + 8 = 0

x= -33 ± √33^2 - 4(4)(8) / 2(4)

x= -33 ± √961 / 8

x= -33 ± 31 / 8

x= -1/4 x= -8 - 24= x^2 - x +30

x^2 - x + 6 = 0

x= -b ± √b^2 - 4ac /2a

x= -(-1) ± √(-)^2 - 4(1)(6) / 2(1)

x= 1± √-23 / 2

x= No Solution