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Week 6 in PreCalc 11

Another week, another blog post for Pre-Calc 11!

 

There were many things that I’ve learned the past week, but one thing we’re going to go over is “completing the square.”

 

First of all, we have to make sure that our QUADRATIC EQUATION (as explained in the previous blog) is EQUAL to ZERO (0).

Now, let’s do an example here. Let’s say that we have…

 

(Unfortunately, latex coding isn’t working well this time…)

^ = to the power of.

 

x^2 + 6x + 8 = 0

Now, what we’re going to do is take the first two terms and make it a perfect square trinomial.

(x^2 + 6x + __) + 8 – __ = 0

So, for this to work, as seen above, we’re going to technically “isolate” the first two terms from the 3rd term and make it a perfect square trinomial. But remember, you can’t just add it like that. Notice that the equation is equal to zero. So this means that you need to have a “zero pair.”

Which means that if you add (let v = ) v from your first two terms, then you must subtract v from the last term.

(x^2 + 6x + 9) + 8 – 9 = 0

So, how did I figure the value for the last term is 9? Here’s the trick.

Divide your middle term by half and square it. So basically if our equation is ax^2 + bx + c, then it will be: (b/2)^2

(x+3)^2 – 1 = 0

So how did we get this? We factor the trinomial (go to my prev blog post for more info) and combined the last two terms.

The next steps should be really straightforward…

(x+3)^2 = 1

Then, just square root both sides. It’ll be:

x+3 = ±√1

x = -3 ±√1

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Week 5 in Precalculus 11

What I learned this week is about quadratic trinomials and how to factor them.

 

A quadratic trinomial looks like this:

ax^2 + bx+ c

Remember that a quadratic trinomial always have a degree of 2.

 

Basically, what you need to do for factoring is find two numbers that will multiply to “c”, and also they must add together to have a sum that’s equal to “b”.

We must not forget signs (- / +) and how they work because they’re also necessary for this unit!

 

E.g.

x^2 + 5x + 6

Find factors of 6:

1 x 6 ; -1 x -6

2 x 3 ; -2 x -3

 

Now, what will add to “5”?

1 + 6 = 7 (NO!)

2 + 3 = 5 (YES!)

 

So now, we’re using 2 and 3 as factors.

x^2 + 5x + 6 =(x + 2) (x + 3)

 

As you’ve noticed, what you did was like doing the FOIL method backwards. For information about FOIL METHOD, check out my blog post about FOIL.

 

 

 

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Week 4 in PreCalc 11

What I’ve learned this week is about adding and subtracting radicals.

But before that, we need to know the parts of a radical.

\sqrt[n]{x} ,where n is the index and x is the radicand.

 

Radicals too, can be added together just like those regular numbers. It’s like adding a fraction, but first you need to make sure your index and your radicand are the same, just like how you need to have the same denominator when adding fractions.

 

REMEMBER: You CANNOT add radicals that DON’T HAVE THE SAME radicands AND index! It’s basically adding “like terms.”

 

E.g.

2 \sqrt{4} + 2 \sqrt{4} = 4\sqrt{2}

 

in words, having 2 copies of 

 

\sqrt{25} + \sqrt{9}

Don’t panic. See if they simplify to a whole number.

\sqrt{25} + \sqrt{9} = 5 + 3 = 8

 

 

Next thing is how to add radicals that don’t have the same radicands.

Note: this might not work for some of them if they have been simplified in their simplest form.

 

The only thing you need to do is to make them have the same radicands by factoring them. It’s based on the index that you have. For example, if your index is two, then you have to find perfect squares such as 4, 9, 16, … and if you have 3 as your index, then find perfect cubes such as 8, 27, … and so on.

For example:

 

\sqrt{27} - \sqrt{3} = \sqrt{9 x 3} - \sqrt{3} = 3\sqrt{3} - \sqrt{3} = 2\sqrt{3}
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Week 3 in Pre-Calc 11

Something I’ve learned this week is about absolute values.

 

Absolute value doesn’t really have a definition. Technically, absolute value is how far away a number from zero is.

 

The symbol for absolute value is: ∣ 

And the value should be inside of those straight lines. E.g. x

 

So, this is how it works. First, let’s look at this number line…

As we can see, “3” is 3 lines away from zero and “-3” is also 3 lines away from zero.

So| 3 | (read as: the absolute value of 3) is 3 and | -3 | is also 3.

Why? Because recalling the technical definition of absolute value, it says “how far” a number is from zero and it’s not asking in which direction it’s from zero. In other words, the absolute value of a number is always positive!

The technical definition should also be like this:

Meaning, the square root and the power of 2 (or square) will cancel each other out.

NOTE: Absolute value doesn’t work like parentheses. | | should not be mistaken as ( ).

For example, -(-2) is NOT -|-2| because:

-(-2) = +2

-|-2| = -(2) = -2

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Week 2 in PreCalc 11

One of the things I learned in Pre-Calc 11 on the second week is about how to add an arithmetic pattern or sequence. I find it awesome because you can know the sum of a whole sequence without adding them all together.

 

An arithmetic sequence is a sequence that is increased or decreased by the same amount of value.

For example, let’s say everything adds up by two(2): 3 → 5 → 7 → 9 → 11 → …

Now, that’s a sequence. What we increase or decrease by is called the common difference (d).

 

But how do we get the sum of these terms without adding them up together?

Before we get to that, when we add the terms of an arithmetic sequence together is called arithmetic series.

 

The difference when you write a sequence and a series is that:

sequence is written like this: 3, 5, 7, 9, 11,…

series is written like this: 3 + 5 + 7 + 9 + 11 + …

 

So how do we add them altogether? Here’s the formula:

S_n = \frac {n}{2} (a + t_n)

 

n is how many your terms in a sequence is.

is your first term.

t_n is your last term.

 

Let’s use my previous example before as the series that we’re gonna use and let’s put the 50th term as our last term.

so it means that … 3 + 5 + 7 + 9 + 11 + …. + 101

That means that our first term is 3, our last term is 101, and the number of our terms is 50.

 

Plug them all in to the formula and remember: PEMDAS.

And we got it! The sum of all terms is 2,600.