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
The first thing we will be learning is Expanding. Essentially we are going to be multiplying binomials and then simplifying.
First we have multiplied the terms.We multiplied the right side with x and then with a.
Now lets try an example with numbers
=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
=x²+5x+6- We added like terms to simplify which gave us x²+5x+6
Now lets try some examples
b) (2x+3) (4x+20
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
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.
Product Sum Method
(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
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+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.
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.
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
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
Now lets try solving some word problems with our new knowledge
To solve this we must set the height to zero and solve for t(time)
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
To solve this question we must complete the square.
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.