Quadratic Lessons Assignment

By Mohamed B.

Introduction

Hello and welcome to my smore. In this, i will be helping you in factoring, expanding, completing the square, and solving. I hope you benefit from this and get a good mark on your test. Enjoy!!

Expanding

Expanding is a form of getting 2 numbers and variable multiplying them (using distributive property) to end up with a standard form quadratic. Expanding is just taking the factored form and putting it back to its original form. Here is an example:


I have a variable and a number in brackets multiplied by another variable and a number in brackets.


(x-12) (x+3)


Now I am going to use distributive property to put it in standard form.


x times x is X2

x times 3 is 3x

-12 times x is -12x

-12 times 3 is -36

If you put it all together, you will get X2 - 9x -36


If you are still stuck, look at the image below. It is another example.

Big image

Factoring

Factoring is basically finding what to multiply to get an expression. In a simple quadratic formula, you have to find 2 numbers that multiply to the c value and what adds to the b value. For example:

Simple Trinomial


I have a equation like this x2 + 5x + 6


So the question is what 2 numbers multiply's to 6 and add to 5. The answer is 3 and 2. This is because 2 times 3 is 6 and 2 plus 3 is 5. Easy right?


Now we have to put it in the expanded form which is like this:


(x+3)(x+2) -- if you expand this you will get the same equation as the one above. It is a cycle!


Decomposition



Now that was the simplest form. There are still 2 more forms to cover. The next one is called decomposition.


If I have the same equation as the one above but instead of nothing in front of the A value, I have a number. This is then when we use decomposition.


So lets say I have this :


2x2 + 5x + 3 - I cant factor this because there is nothing in common. Therefore, I will multiply the A value by the C value and do the same thing as before; which is to find 2 numbers that multiply to the (new value) and add to the B value.


ok so 2 times 3 is 6. What multiply's to 6 and adds to 5. 3 and 2. Now we put it like this:


2x2 +3x +2x +3 - what you do is that you remove the b value and put the 2 new numbers in and add an x beside it.


Now we have to factor by grouping. This is a sub category. Basically you take the first 2 numbers/variables and you factor and then you take the next two varialbes/numbers and you factor.


3/2x(4/3x+2)+3/2(4/3x+2)


Then if you join the things that are not in brackets together and take out 1 of the same things in brackets, you will get this:


(3/2x+2)(4/3x+2)


And there you go. By the way sorry, I had to use fractions. Usually fractions/decimals are not used in decomposition.


Difference Of Squares


The last factoring thing is called difference of squares.


Ok so lets say I have an equation like this


x2 - 49y2


If all these can be square rooted and there is a negative sign between them, we can find the difference of squares.


This is what you do.


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


We get the root of it and then we do x + whatever the number and x - whatever the number. Also it can have more than 1 variable, different variable, etc. (not necessarily x.)



Still having trouble with difference of squares? Watch This !!

Big image

Completing The Square

Completing the square is simply converting from standard form to vertex form. It is a very simple process and gives you useful information such as the stretch/compress and the vertex. Do do this there are a few steps. Lets use the equation 4x2+ 2x+6 as a example.


1. If there is a A value factor it out. In this case their is


4(x2+0.5x + 1.5)


2. Take the C value out but multiply it by the factored A


4(x2+0.5x)+6


3. Divide the B value by 2 and square it. Then add the answer in the equation.


4(x2+0.5x + 0.0625)+6


4. The same number that you divided by 2 and squared, make it a negative and multiply it by the A. Then place it outside the bracket


4(x2+0.5x + 0.0625)+6 - 0.25 ( subtract 6 - .25)

= 4(x2+0.5x + 0.0625)+5.75


5. Now see that value that was divided and squared and added in the brackets. Get rid of it.


= 4(x2+0.5x)+5.75


6. And the last step is to divide the 0.5 or B value by 2 and square the whole bracket. Also get rid of the square on the x.


4(x+0.25)^2+5.75


And there you go. You have completed the square successfully!


Still stuck, this video should help.

❤² How to Solve By Completing the Square (mathbff)

Solving a Quadratic Equation

Solving the Quadratic Equation is very simply. Its is pretty much finding the 2 x intercepts on the graph. There are 2 ways of doing this. To do it, you have to factor. Here is an example. I have it in standard form.


1rst way


Standard form

x^2 -2x-35 = 0


Now what 2 numbers multiply to -35 and add to -2. -7 and 5 right?


(x-7)(x+5) = 0


Now we have to solve for x. So what 2 numbers make 0. The first one is x = 7 and the second one is -5. If you put those variables in for the equation it will be 0. Here is how.


(7-7)(-5+5) = 0 ---- this equation makes sense. So the final answer is x = 7 and x = -5.


The 2nd way of solving is say I end up with something like this.


(2x-5)(2x+3) = 0


I have to solve for x right. So I will move the number to the other side and divide by the variable. You will end up with this.


x = 5/2 and x = -3/2


Still having trouble. This last video will help.