Quadratics Review Assignment-Unit 4

By: Zoya Naseem

Linear and Non-Liner relations

The differences between linear relations, non-linear relations and quadratic 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

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Quadratic Formula

Quadratic formula is used when finding the x value using an equation

ax^2+bx+c

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

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Multiplying Binomials

To multiply binomials, it is important to use the distributive property to expand the binomials and collect those like terms in order to simplify

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

Common factoring is the opposite of expanding, in order to factor a polynomial, it is important to find the greatest common factor (GCF).


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.

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Factor by Grouping

Factor by grouping is when you make an expanded form into binomial. You can expand it by using common factor depending on the question, or use sum and product method if common factoring is not possible.


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

Factoring simple trinomials is when a trinomial is converted back to two binomials also to make binomials, we would use sum and product method. Factoring simple trinomial relates to factor by grouping. This is known as type 1.


=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

The difference between factoring 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

Solving quadratics by factoring is almost the same as factoring trinomials but in this case we find the "x" values by making the expression into an equation.


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

Completing the square is when you make the expression, ax^2+bx+c into this equation

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


Example: ---------------------->y=x^2+8x-3

  1. Block the first two terms ------------> y=(x^2+8x)-3
  2. Factor out the "a"
  3. divide the middle term by two, then square it -------------------------------------->(8/2)^2=16 y=(x^2+8x+16-16)-3
  4. take out the negative number inside the bracket-----------------------> y=(x^2+8x+16)-16-3
  5. 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