In week 14 of precalc 11 we went even deeper into fractions and finding non permissable values, this time I will talk about multiplying and dividing fractions. When multiplying you can simply just multiply across then maybe divide to simplify, then you will have your answer lets use this example 5/3x+2 * 2/x+1 =?. We literally can just multiply numerator by numerator and denominator by denominator. so this will give us 10/3x+2. This isnt really factorable further so that will be our answer. Now to divide fractions it’s also pretty simple. The secret is that you can just multiply the reciprocal of the second fraction. Let’s use a complicated fraction such as this one.

It looks pretty scary but it’s easier when you write it horizontally. Anyways we start by simply turning the second on into it’s reciprocal and multiply it

Just before we multiply it, luckily this one can be simplified first. we simplify by finding multiples in the diagonal spots of the fractions. This means the left numerator and the right denominator and vice versa. The 3x and 6x can be divided by 3x, resulting in 1 and 2. The left numerator and right denominator can be factored resulting in (x+3) (x+2) on the left and (x+7) (x+2) on the right. This is what it will look like when we simplify it.

next we can divide the (x+2) on the left numerator and right denominator. After that we multiply (but leave it in factored form from here) resulting in 2(x+3)/(x+7) and since we can’t factor further, that is our answer and the non-permissible values are 0,-7 ,-2 because if those numbers were x, it would end up in the denominator being 0 and thus the fraction doesn’t work.

In week 13 of precalc 11 we dove deeper into the whole non-permissible values thing. Somehow, Mr and Mrs. Math found a way to make it even more complicated. We now have more ways to make the denominator = 0. We can do so by adding or subtracting fractions with each other.

In last weeks reflection I forgot to mention that if the denominator is just X, you can simply write X≠0 because that works.

Now to add fractions to find the non-permissible values, lets start with a fraction like 2/x+3 + 1/x-2. Firstly we know our restrictive values can be -3 and 2, but lets do the equation so we can really find what they are. We can multiply both denominators and make the fractions have the same denominator. This will result in 2x-2/x^2-6 +x+3/x^2-6. we then add the numerators, which will result in 3x-1/X^2-6. the denominator can’t be zero and to make it 0 it will have to end up being 6-6, so x can’t be the root of 6 because x^2 will result in being 6. this means the non-permissable valv=ues are 2,-3 and √ 6. This is how you add fractions and find non-permissible values

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This week we reviewed inequalities. This means the signs including <,>,≥,≤ and  ≠. We learned how to graph using them (with linear equations) and how they affect the graph. No mater the sign (not sure about ≠), you will end up shading over (>≥) or under (≤<) the line which will either be regular(≥≤) or dotted(><). We do this because anything over (>≥) or under (≤<) the y value is free game because it’s still true.

Let’s use a couple of simple equations as examples:

y>3x+7: Dotted line and anything over the Y value,

y≤2/5x-3: Regular line and anything under the Y value

In week 8 of precalc, we went over review on the quadratic formula, then went over discriminant, then we went over graphing. A quick definition of the discriminant is the

√b^2 -4ac part of the quadratic formula. The point of a discriminant is that it will help you know how many answers you will get. If the discriminant is over zero you will get 2 different answers. An x1 and an x2. If it is zero then you will get only 1 answer. If it is less than zero you will get no answers because you can’t root a negative. Basically, finding the discriminant will help you know if there is a solution and how many in a fast way.

On to graphing, graphing is relatively easy. Making the T chart is the hard part. In a standard graph there are X and f(x)/Y values. The X values are known as the input and f(X) or Y is the output. So far, we have learned about quadratic functions (Y=X^2) and linear functions(Y=X). We have been doing equations in which we find f(x) or Y to see how the input and output values work. Lets use this quadratic function equation as an example.

f(X)=-X^2

This might seem simple but a mistake I made was forgetting the order of operations and making f(X) positive as I thought it was a trick question and a negative number squared is positive. I was supposed to be squaring, then making it a negative.

Now with this equation we start by making a T chart

After making the T chart, we put dots on the graph in the position indicated by the numbers we have, such as (X3,Y-9) and (X0,Y0) and so on. After putting in the dots, we connect the dots with a straight line, for this type of equation, the lines will usually be drawn going out of the chart. Furthermore, for this specific equation I decided to leave in the parent equation for quadratic functions (in blue) with the equation we used as an example (in red) to see what has changed.

That is pretty much the basics about graphing quadratic functions and what a discriminant is.

on week seven of precalc we learned how to do quadratic equations with the quadratic formula. The quadratic formula is probably supposed to be the last bastion after trying everything else when it comes to solving equations. It’s like the last thing you’d want to use, while reliable, it takes a minute to set up. It was made to always work and is a lot easier with a calculator as you just plug numbers in. The formula only works problems with the formula Ax²+Bx+C=0 and looks like this

We also watched a song to the tune of pop goes the weasel and that makes it easy remember.

Now, to use the quadratic formula, you need to know where everything is supposed to be. A will replace the x², B is the coefficient to the ‘x’ and C is the remaining constant.

Let’s consider a quadratic equation:

Here, A=2x, B$4$, and C$6$. Using the quadratic formula, we substitute these values:

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So, the solutions to the quadratic equation 2x²+4x−6=0 are x1=1 and $-3$.

In week √36 of precal we learned about quadratic equations. Quadratic equations are unique in the way that they will always be AX^2 +BX + C = 0. An interesting thing about quadratics is that there are different ways of solving them. You can factor them or you can “complete the square”. to complete the square you move the number in the C spot over, find the number that would make it a perfect square, then add it to both sides and simplify further from there. I will go over how to do that, including how to find what number will make it a perfect square.

Lets use the equation X^2+6x+8=0 as an example. To start, we subtract the 8 from the left side and the right side

X^2+6x=-8

Next we find what number goes into the C spot. In this certain equation when there is no coefficient to X^2, it will be half of the coefficient of B squared. In this it will be 3^2 which is 9. We then add 9 to both sides and it will look like this

X^2+6x+9=1

Now we can easily factor the left side to simplify the equation.
(x+3)^2=1

Now we must remove the exponent on the left side and we can do so by rooting both sides

√(x+3)^2=√1 = X+3 = +/-1

because of how rooting works the 1 is either positive or negative, meaning there are 2 answers to these types of equations.

X1=-2    X2=-4

And this is how you solve quadratic equations by “completing the square”.

In week 5 of precal we learned how to factor equations with 4 terms. Using the box method makes doing these things a breeze. Let’s use the equation X^2+6x+2x+12 as an example. we start by drawing a box in the middle and putting the greatest term (usually x^2) at the top left and leave the shortest term (usually a number with no variables) at the bottom right with the terms in the middle (usually variable with no exponent/1) in the other corners. I like to add a positive sign or negative sign on all of the numbers so I don’t make any mistakes.

Now that we have all of our terms in a box, we have to find what goes into all of these terms and place them outside of the box like so:

Now that we have found what goes into the terms we re-write the equation. for this equation it will be (X+6) (X+2) and that is completely factored.

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This week of precal we learned about division of radicals and rationalizing the denominator. It is (obviously) very helpful to rationalize the denominator in a question where the denominator isn’t a rational number. Let’s use this equation as an example:

Now, we know that the denominator is irrational. In order to solve the equation we must rationalize it and the easiest way to do that is by multiplying it by the conjugate of itself (√5+1), remembering that what you do to the bottom you must do to the top.

This is what you get from multiplying the top and bottom by the conjugate of the denominator (√5+1).

In this particular equation we can further simplify it because there is a common factor (4). we can divide the 4s on the top by the 4 on the bottom resulting in √[5+4]. And that is how you simplify a division equation that has an irrational denominator