# Quadratics Review Assignment-Unit 4

### By: Zoya Naseem

## Linear and Non-Liner relations

__Linear Relations__is when a straight line can be drawn through all of the points.

__Non- Linear Relations__is when a table of values represents a single smooth curve that can be drawn through every point.

__Quadratic Relations__is when table of values represent a constant increments of the independent (x), the second differences of the dependent variable (y) are constant.

**(Δy**

**) means first differences.**

__Degree__

5x has a degree of 1

x^3 has a degree of 3

2x^2 has a degree of 2

5x^2-3x+2 has a degree of 2

## Quadratic Formula

ax^2+bx+c

The video below will explain more and give examples on quadratic formula.

## Multiplying Binomials

__Steps to solve:-__--------->(x+a)(x+b)

1st step is to multiply the first bracket with the second bracket

(x)(x)=x^2

(x)(b)=bx

2nd step is to multiply "a" with the rest in the other bracket

(a)(x)=ax

(a)(b)=ab

3rd step is to expand it

=x^2+bx+ax+ab

=x^2+ax+bx+ab

=x^2+(a+b)x+ab

How I got this:

x^2 because (x)(x) gives x^2

(a+b)x because you add the like terms

ab because a value times b value gives us the number

__Example__

(x+2)(x+3)

=x^2+3x+2x+6

=x^2+5x+6

## Common Factoring

xy+xz=x(y+z)-----------------------same as------------------------> 5x+35=5(x+7)

I have made 5x+35 into 5(x+7) because 5 goes into 35, 7 times.

## Factor by Grouping

__Trinomial__= **(Binomial)(Binomial) ----------------> Quadratic Expression**

x^2+15x+56

Sum:15 and Product:56 ---------------->(7)(8)=56

--------------------------------------------------->7+8=15

(x+7)(x+8)

The video below will give more explanation and examples about factor by grouping

## Factoring Simple Trinomials

=x^2+7x+6

Your b (what it is) should be the sum of two numbers and those 2 numbers should be the product of c value. And in this case the two numbers are.....

Sum: 7 Product: 6----------->(6)(1)=6

-------------------------------------> 6+1=7

=x^2+6x+1x+6

=x(x+6)+1(x+6)

=(x+1)(x+6)

## Factoring Complex Trinomials

__simple__trinomials and factoring

__complex__trinomials is that for simple trinomials, we would have to just find the sum and product but in complex trinomials, the "a" value would be given so we would have to multiply the "a" value with "c" value and then find the sum and product. This is known as type 2.

ax^2+bx+c------------------> Expression

__Example:__

= 8x^2+22x+15

=(8)(15)=120

=8x^2+22x+120

Sum: 22 and Product:120----------->(10)(12)=120

----------------------------------------------->10+12=22

=8x^2+10x+12x+15

=2x(4x+5)+3(4x+5)

=(4x+5)(2x+3)

## Solving Quadratics by Factoring

__Example 1: __

2x+5=0

2x/2=-5/2

x=-5/2

__Example 2: __

x^2-5x-14=0

Sum: -5 and Product: -14----------->(-7)(2)=-14

-----------------------------------------------> -7+2=-5

x^2-7x+2x-14=0

x(x-7)+2(x-7)

(x+2)(x-7)

x=-2, x=7

**Note:** any numbers that move across the equal sign, changes into the opposite sign

## Completing the Square

y=(x-h)^2+k which is known as completing the square/vertex form.

__Example:__ ---------------------->**y=x^2+8x-3**

- Block the first two terms ------------> y=(x^2+8x)-3
- Factor out the "a"
- divide the middle term by two, then square it -------------------------------------->(8/2)^2=16 y=(x^2+8x+16-16)-3
- take out the negative number inside the bracket-----------------------> y=(x^2+8x+16)-16-3
- y=(x+4)^2-19 <---------------------------This is known as vertex form

__Note: __ If the "a" value had a number then on the 4th step, multiply the "a" value with the negative number which is inside the bracket, then take it out of the bracket.

## Challenging Questions

__Question 1:__

**2x+1=√2x+1**

(2x+1)^2=(√2x+1)

(2x+1)^2=2x+1

[(2x+1)(2x+1)]-(2x+1)=0

[4x^2+2x+2x+1]-(2x-1)=0

(4x^2+4x+1)-2x-1=0

4x^2+2x+0=0

2x(2x+1)+0=0

2x=0, 2x+1=0

x=0,x=-1/2

__Question 2:__

**n^2-n(n-1)/2=0**

(2)n^2=n(n-1) <----- cancelling the 2 out to get rid of the 2 that was being divided by n(n-1)

2n^2=n^2-n

2n^2-n^2+n=0

n(n+1)=0

n=0, n+1=0

n=0, n=-1

__Question 3:__

**(x-3)x=4**

x^2-3x-4=0

sum: -3 and Product: -4-------->(-4)(1)=-4

------------------------------------------>-4+1=-3

x^2-4x+1x-4=0

x(x-4)+1(x-4)=0

x+1=0, x-4=0

x=-1, x=4