Quadratic Word Problems

Algebraic Portion

Intro

In this review package we will be covering the algebraic portion of the quadratic unit. In this unit we will be covering Expanding, Factoring, solving and completing the square.

How to expand

Expanding in a quadratic brackets we are expanding them to get a quadratic equation in standard form. Lets look at some examples and learn how to solve them.
Example1: (x+2)(x+3)

Now to expand this we are going to be using the foil method. Meaning First, Outer, Inner and last.

So the first step in the foil method is multilying the first digits in each brackets. In this case we multiply x from the first bracket with the x in the secind bracket. and put our answers on the right side of the equation.

(X+2)(x+3)=x^2

Doing that gave us a x squared as the first part of our factored equation. The second part step in the foil method is called Outer. Here we multiply the outer digits with each other. In this case its x and 3.

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

Doing that gave us a answer of 3x and the second part of our equation.

The third step in the foil method is called inner. Here we multiply the inner digits with each other which in this case are 2 and x.

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

doing that gave us a answer of 2x.

The final step in the foil method is called Last. In this step we multiply the last digits with each other, which in this case are 2 and 3.

(x+2)+(x+3)=x^2+3x+2x+6

Doing that got us 6 as our last digit.

So we got our expanded equation as x^2+3x+2x+6

but we have some like terms. So our next step is to add up the like terms and add them up. In our equation we have 3x and 2x as the light terms so we add them up and get a simpler equation.

x^2+5x+6

Thats it where done. Refer to the video down below if you are still not clear on the concept

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PSAT Math: Using FOIL to Solve Quadratic Equations

Factoring and solving Quadratic Equations

When we are factoring we are doing the opposite of expanding. We are working back words and converting it into two polynomials. Lets work through a example.

x^2-5x-6

when factoring this problem we first have to identify our a,b and c.which are

a=x^2

b=-5

c=-6

Now we look at the b and c and figure out what what multiplies to 6 and adds to 5. This may take a while. Use guess and check if you still cant figure it out we will look at some methods that you can use if you are unable to find your numbers. In this case the numbers we are looking for are -6 and positive one. Next we setup our two polynomials.

We start of with x in the brackets.

(x ) (x )

Now we add our two numbers we found -6 and -1 to the polynomials. And it becomes

(x-6) (x-1). And now we have successfully factored our equation. But where not done we still have to slove for x. What we do is open up our brackets and set them to equal zero.

x-6=0 and x-1=0 and now using our basic algebra skills we can solve for x.

x=6 or x=1

Now we end up with two solutions x=6 or x=1. Thats it we solved it.

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❤² How to Solve Quadratic Equations By Factoring (mathbff)

Grouping Factoring Method

The grouping factoring method is another method used for when you want to factor a equation. Lets look an example.

12x^2+34x+10

Now what we first want to do is factor out the greatest common factor which in this case is 2. So our equation will become.

2(6x^2+17x+5)

Now lets identify our a,b and c terms.

a=6

b=17

c=5

What we do is multiply our a term with our c term in this problem we get 30. Now we have to find two numbers that multiply to 30 and add up to 17. 2 and 15.

Now we rewrite the equation replacing our b term with our two new numbers.

2(6x^2+2x+15+5)

Now separate the first two numbers with the second two and find the greatest common factor for both of them. It should be the same.

In this case we get (3x+1).

Now we take the numbers that we used to find the greatest common facor and put them in one bracket and take the greatest common factor and put it in another.

(2x+5) (3x+1)

SInce in the beggining we used a 2 to find the common factor of the equation we have to stick it in the front,

[2(2x+5)(3x+5)]

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solving complex equations

Now lets solve a complex quadratic equation with a number in front of the coefficient. Lets look at how to solve this.

6x^2-5x-4.

Our first step like the last one is to identify our a,b and c. In this case they are

a=6

b=-5

c=-4

In a complex trinomal we also have something called ac thats when we multiply our a with our c. So 6*-4=-24 so our ac value is -24

Now we have to find the a number that multiplys to -24 and adds -5. In this case the numbers are -8 and 3. Now we put these into the equations like this

6x^2+(3-8)-4

6x^2+3x-8x-4

Now we use our grouping factoring method and get.

(6x^2+3) (-8x-4)

3x(2x+1)-4(2x+1)

(3x-3) (2x+1)

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Factor a Complex Trinomial

Quadratic Equation

Quadratic equations is a way to slove a quadratic formula that you can not find a number that will add up to your b value and multiply to your c. Lets take a look.

x^2-5x-6=0

Lets say you cant find a number that will add up to -3 and multiply to -4.

We first have to indifinty our a,b and c values.

a=1

b=-5

c=-6

Then we use the quadratic formula that looks like this.

x=-b+-√b^2-4ac/2

Lets plug our equation into this formula.

x=5+-√-5^2-4*1*-6/2a

we can simplify that into

x=5+-√25+24/2

Simplify that even more

x=5+-√49/2

and now we end up with two solutions

5+7/2=12/2=6 5-7/2=-2/2=-1

Thats is that how we solve it using the quadratic formula.

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Using the Quadratic Formula

Complete the square

Completeing the square is another way to slove quadratic equations.

x^2+6x-7=0

Now our first step is to move the last constant to the rite side of the equation which will look like this

x^2+6x=7

Our next step is to add a number to both sides. We get that number by dividing the middle constant by 2 and squaring it.

6/2^=9

So we got nine. We put in in the equation on both sides.

x^2+6x+9=7+9

We can simplify the equation to

x^2+6x+9=16

Now we can write it as a perfect square.

(x+3)^2=16

Now we square root both sides.

√(x+3)^2=√16

x+3=4

since we square rooted it we get two solutions one that four is a negative number and one that it is a positive number.

x+3=4 x+3=-4

x=4-3 x=-4-3

x=1 x=-7

So our 2 solutions are x=1 and x=-7

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