This week in Precalculus 11 we learned about discriminant. A discriminant is a part of the quadratic equation that helps us determine many things such as if the equation is factorable, how many roots it has, or if it has any roots at all before we actually solve the whole thing.

Example

Equation:

Step 1: make sure to take out anything common

Step 2: determine what a, b, and c are.

a = 1

b = -6

c  = 9

Step 3: fill the formula

Step 4: simplify

(36) – (36)

0

Explanation

Step 1: make sure to take out anything common

Taking out things that are common makes figuring out the answer a lot easier because the numbers are smaller to work with, but even if you leave everything in, it comes out with the same answer.

Step 2: determine what a, b, and c are

The formula for finding the discriminant relies on a, b, and c, and they are all in coordination with the equation. a would be equal to the first number (the number that has attached to it. b would be equal to the middle number (the number attached to x). c would be equal to the last number (the number that has no variable attached to it.). Writing out what a, b, and c are equal to makes it a less chance of error.

Step 3: fill the formula

Next, fill the formula according to coordination between letter and number

Step 4: simplify

There’s not much left to do except to find out the answer. The answer for this example was 0. This indicates that this equation has one real root. If the answer was any number bigger than 0 then that would mean that this equation has 2 real roots. If the answer was a number smaller than 0 (any negative number) then the equation has no real roots which means there is no need to solve the rest of the quadratic equation. That’s why it is helpful to find the discriminant, because we don’t need to solve more than we need to if we don’t need to.

This week in Pre-Calculus 11 we reviewed the different ways to factor, we learned how to find out what x can equal in a quadratic equation when we can factor it, and we learned how to find out what x can equal when a quadratic equation isn’t factorable.  The ways we can do that is by completing the square or filling in the quadratic equation. The most difficult thing I think we learned is completing the square.

Completing the Square is a method used to solve a quadratic equation by changing the form of the equation so that one side is a perfect square trinomial. This makes the equation ‘kind of’ factorable and able to solve for x. Without the completing the square method, there would be no way to solve for x for a non factorable trinomial.

Example:

Explanation:

Step 1- find out if it’s a quadratic equation and if it’s factorable

To find out if it’s a quadratic equation, look for an exponent in the equation that has a degree of 2. To check if it’s factorable, multiply the coefficient attached to x2 and the number with no variable attached to it. In this case it would be 26. Now, I would check if two numbers that multiply two 26 also add up to the middle term then it’s factorable. In this case, the equation is non factorable, but it is quadratic.

Step 2- make sure x2 has a coefficient of one

If it doesn’t have a coefficient of one then make sure the number can be divided by the middle term. If it does, then divide the x2 coefficient and the middle term coefficient by the x2 coefficient to make it a 1.

Step 3- divide the middle term by 2 and square it.

At this point, we are at an equation that cannot be solved for x. We can make it solvable by completing the square. The middle term in this example is 6. So if we divide that by 2, which is 3, and square that, which is 9, then we just completed the square.

Step 4- add the # to the equation with a zero pair.

So if we add 9 to the equation (after the 6x), then we have to make it a zero pair by adding a negative 9. If we don’t add the negative 9 then the whole equation is changed, but adding a negative 9 makes a zero pair so it basically cancels each other out, but is very significant.

Step 5- multiply the 2 with the 9

After adding the zero pair, then brackets go around the first three terms. In this example it’s 2(x2 – 6x +9), and the negative 9 goes on the outside of it with the +13. In this case, the 2 on the outside of the bracket is multiplying the positive 9 that we added. To keep the zero pair we have to multiply the negative 9 by the 2 as well.

Step 6- factor and solve

The reason the brackets went around the terms before is because it factors perfectly into two factors that are exactly the same. Here it’s (x-3)(x-3). To put it in simpler terms its (x-3) squared. Now there’s not much left to do except to solve for x.

First move the -5 to the other side of the equal sign. Now to get rid of the two, divide both sides by it. Then to get rid of the square, square root both sides. When you add a square root sign to the fraction, you must put a ± sign before it because the outcome of the square root can be either a negative or a positive. Last, move the three over to the side with the fraction.

Step 7- don’t leave the denominator as a radical

As we learned last unit, radicals can’t be left in the denominator. So to get rid of it, multiply the top and bottom of the fraction by the square root of the denominator which is two. This makes only the top a radical now because the denominator will be equal square root of 4 which is 2.

Step 8- simplify where necessary.

This example didn’t need any because 10 can’t be broken broken down anymore.

This week in PreCalc 11 we learned how to solve radical equations (which is an equation where the variable is under a radical). It was easy if we used our basic knowledge of equations; what we do to one side we do to the other (if I added 25 to one side I would add 25 to the other side of the equal sign)  AND to cancel out an operation you do the opposite of it. (to cancel out -25 I would +25). With this basic knowledge it was pretty straight forward.

Example:

Explanation:

Step 1: apply the opposite operation to get rid of the square root

In the example, square roots were surrounding all the numbers in the equation so not much could be done without them gone. To get rid of the square roots you have to find the opposite operation which in this case will be to square everything. This is because once we square everything and it stays under the square root, the square root will cancel it out.

Step 2: bring like terms together

Now that the square root isn’t in the way of solving the equation we can do it now. First thing’s first, bring like terms together. The 5x and 2x are on opposite sides. So to remove the the 2x from the side it started on originally we have to subtract 2x to cancel it out. Then, since what we do on one side we do to the other, we have to add 2x to the other side. Now the -1 and 5 are on different sides of the equal sign so we have to +1 it started on originally and then +1 on the other side. After all the like terms are on the same side, combine them.

Step 3: SOLVE

All there really is to do is solve. To isolate x we have to divide by 3 and do that to the other side too. Then we are left with the answer which is 2.

Step 4:

Now that we know that x = 2 we need to verify our solution to make sure the root of the equation is a real root. For it to be a real root x has to be greater than or equal to 0. And once we have determined that, we have to put what x equals back into the original equation. In this case when I put 2 back into the original equation, both sides equal 3, therefore; 2 is a real root.

This week in PreCalc, we expanded on the ideas of radicals by learning how to add, subtract, multiply, and divide them. A lot of the ideas were straight forward, but the most challenging is dividing radicals.

Example:

Work:

Step 1: simply if necessary

Simplifying the given equation if it can be simplified makes the work easier later on because the numbers will be smaller.

Step 2: identify the conjugate

For dividing by a binomial, you have to multiply the numerator and the denominator by the denominators conjugate (is formed by changing the sign between two terms in a binomial). So the binomial in my example has a negative sign, but it’s conjugate will have a positive sign, and that’s the only difference between conjugates. The rest of the terms stay the same.

Step 3: multiply top and bottom by conjugate

Start with expanding the top and the bottom.

Step 5: simplify where necessary

After expanding, combine like terms, simplify radicals, and solve until you no longer can. When you can’t go any further, then you have reached your answer.

This week in PreCalc 11, we learned about Absolute Values (of a real number, is the square root of the square root of a number. In simpler form, it’s the distance of a number on the number line from 0, and can only be a positive. So, the absolute value of -5 would be 5) and Roots/Radicals (a radical expression is defined as any expression containing a radical (√) symbol. It can be used to describe a cube root, a fourth root, or higher depending on the what the index is. The index is the number on the top left corner outside the radical symbol. A root is, when multiplied by itself is given a number. So, 2 is the root of 4 because 2×2=4)

The things we did were pretty straight forward this week, but the hardest thing I came across was evaluating expressions that had absolute values in them.

Example:

Explanation:

Step 1: find the absolute value first

I personally do this first because when the number you’re trying to figure out the absolute value for is a negative, then it turns into a positive so it’s easier to get that out of the way first. In this example, the value of a positive so we just keep it a positive

Step 2: finish the rest of the expression

The rest is simple BEDMAS, and trying to find the answer. Don’t forget, the math in the brackets should be done first, and then the rest follows.

This week in Precalc 11 we learned about Geometric Sequences (patterns that multiply by the same number to get the next number), Finite Geometric Series (patterns that don’t have numbers that eventually start to look the same), and Infinite Geometric Sequences (patterns that never have an ending, but still possible to find the sum because eventually the numbers in the sequence become very close together). The one thing that stood out to me is finding the finite geometric series.

When given a geometric sequence, just like an arithmetic sequence, it’s possible to find the sum of it. Finding it for an arithmetic sequence, though, is different than finding it for geometric sequence. For example, if you need to find of a geometric series you don’t need to find first like for an arithmetic series. Instead, you can just plug in what you know into the equation given.

Ex.

Explanation:

Step 1: Find the common ratio

Common ratio is different than the common difference that we were introduced to in the arithmetic unit. Instead of finding the number that the numbers go up by each time, we find the number that is used to multiply the last number to get the next number.

Step 2: Find

Once we know what the common ratio is, then we have everything we need to fill in the equation to get the answer we are looking for.

This week in Math 11 we learned about Series and Sequences. Sequence is essentially to just a pattern that starts at a number and continually goes up my the same number each time. And a Series is the sum of all the numbers of the sequence . The one thing that stood out for me and was challenging to learn was figuring out how to find being given and . After figuring it out, it’s safe to say it’s essentially just figuring out what you don’t know, what you do know, and which equation to use.

Example:

The Work:

Step 1: Figure out what you know and don’t know based on what’s given in the question

In this example, we know and and we don’t know what or is equal to or what is equal to.

** is the first number in the sequence the series represents

** or or any number that is attached to the is the general term that represents the place of whatever the number is ( is the 15th term in the sequence)

**  is the difference between the number add uns in the pattern which has to stay the same throughout to be a artithmetic sequence

Step 2: Find d

Finding D is the most important in this question because before we can find we have to find and we can’t do that without knowing the difference between each term in the pattern

Step 3: find

Finding the 30th term is important too because of its connection to which is that for there to be an there has to be a 30th term in the sequence that the series can add up until. And to figure out we need to know what is because we need to know what the number is to add in the series

Step 4: Figure out what you know now and what you still need to know

In this example, now we know at this point that , , and  and we still don’t know what is equal to

Step 5: Find

At this point, we have all the information we need to use the series equation and find what we were looking for in the beginning