Week 4 – Simplifying Radical Expressions

Posted on by shannonf2015

I learned how to expand and simplify a radical expression.

(3\sqrt{5} + 7\sqrt{8})(\sqrt{10} – 6\sqrt{5})

There are multiply ways of starting to simplify this expression, but the first thing that I do is see if I can simplify the radicals given.

Looking at the expression, I can see that I can simplify 7\sqrt{8}, as 8 has a square root, which is 4.

7\sqrt{8} —>  7\sqrt{{4}\times{2}} —> 7(2)\sqrt{2} —> 14\sqrt{2}

The next step is to distribute the values in the first bracket and multiply with all the values in the second bracket.

This process is called FOILING.

F – first terms

O – outer terms

I – Inner terms

L – left over terms

Rule: We can simply multiply the radicals together, unlike adding and subtracting where they have to have the same index and radicand.

So in my example, it doesn’t simplify by a lot as they don’t have any like terms.

The answer would therefore be :  -90 + 15\sqrt{2} + 28\sqrt{5} – 84\sqrt{10}

Since my example isn’t the best to show how to simplify when it comes to like terms, let’s look at this example.

I did this equation in two ways which does give you the same answer. The first way, I simplified the radicals first, then added the like terms together. Note: They have to have the same radicand and same index in order to add them or subtract them together.

The second method, I just added the like terms together first as 4\sqrt{x^3} had the same radicand and index as -7\sqrt{x^3} and when you simplify 2\sqrt{x^2}, it becomes 2x which can be added with 3x. So when you add the coefficients together keeping the same radicand and index, you are combining like terms.

Then, I simplified the radicals, {-3}\sqrt{x^3} can be simplified to {-3x}\sqrt{x}.

Since we have a variable in our expression, we must define x. Our index here is 2 (square root), therefore, x must be greater than or equal to 0 because we cannot have two numbers that are the same that will multiply to give a negative number: x ≥ 0

Last thing that I have learned was simplifying fractions with a radical as the denominator.

Since we cannot leave radicals in the denominator, we must rationalize the denominator in order to make it a real number.

Rationalizing the denominator means multiplying both the nominator and denominator by the denominator. I included the coefficient but you don’t have to. It would probably be best to just multiply \sqrt{5} with the denominator and nominator, but both give the same results.

After I multiplied them, I noticed I could simplify even more. Then finally, I simplified the coefficients as both can be divide by 10.

 

 

 

Week 3 – Roots and Radicals

Posted on by shannonf2015

I have learned how to simplify radical expressions, as well as changing mixed radicals to entire radicals.

Part 1: How to simplify a radical

\sqrt[3]{135}

As we can see, 135 is not a perfect cube root, but we can simplify this radical by finding a cube root that divides into 135.

First let’s look at some vocabulary :

      

We can see that the index in this example is 3. So the first step is looking through the list that corresponds with the index. In this case it is cube roots.

List of cube roots:

\sqrt[3]{1} = 1

\sqrt[3]{8} = 2

\sqrt[3]{27} = 3

\sqrt[3]{64} = 4

\sqrt[3]{125} = 5

\sqrt[3]{216} = 6

Starting from the top of the list, we can tell that 1 is not going to help, so we move on to the next number which is 8, but that is not a factor of 135 as it doesn’t divide in the number evenly without resulting in a decimal. So we go to the next cube root on the list, which is 27.

27 does divide evenly into 135 which equals 5.

So the radical would look like this :

We can now simplify it by determining the cube root of 27.

\sqrt[3]{27} = 3

That number now becomes the coefficient of the equation and the 5 remains as the radicand :

\sqrt[3]{135} = 3\sqrt[3]{5}

** Don’t forget to always include the index **

Part 2: How to write a mixed radical as an entire radical.

2\sqrt[4]{7}

In this example, we can see that the index is 4, the coefficient is 2 and the radicand is 7.

To be able to turn this into an entire radical, we must be able to put the coefficient (2) back into the root while dragging the index in too.

So we must determine what is 2^4 and then place that number into the root and multiply both numbers while keeping the index:

 

Therefore, 2\sqrt[4]{7} = \sqrt[4]{112}

 

 

 

Week 2 – Geometric Sequences

Posted on by shannonf2015

This past week, I have learned many different formulas and techniques to solve geometric sequences such as finding the sum, a term, and common difference. One of the most important concepts I have learned was how to find a term when given two terms that are not consecutive.

My geometric sequence problem:

In a geometric sequence, the third term is 189 and the sixth term is 5103. Find the eighth term.

Terms:

r = common ratio

a = first term

Step 1: Write down the given values.

t_3 = 189

t_6 = 5103

t_8 = ?

Step 2: Decide which equation would work best with the information given and what info we are missing.

= t_n={ar^{n-1}}

As we can see in this equation, we are missing a=t_1 and r=common ratio.

To find these values, we must first start by finding r to be able to find the first term.

Since we are given the values for two terms: t_3 and t_6, we can find the relation between them in their geometric sequence to find r.

In this image, it shows that to start from the third term, it takes three multiples of the common ratio to get to the sixth term. The equation to find r would look like this:

= t_3\times{r}^3=t_6

Now plug in the values given: t_3 = 189, t_6 = 5103

= {189}\times{r}^3=5103

To find r, we must isolate {r}^3. To isolate {r}^3, you must divide 189 on both sides of the equation.

\frac{189}{189}\times{r}^3={5103}\div{189}

{r}^3 = 27

\sqrt[3]{r} = \sqrt[3]{27}

r = 3

Step 3: Now that we have the value of r, we can find t_1.

Starting from t_3 to go to t_1, we must divide by how many times they are spaced out by the common ratio, as it’s going backwards.

Equation would look like this: t_3\div{r}^2 = t_1

189\div{3}^2 = t_1

189\div{9} = t_1

21 = t_1

Now that we have found t_1 and r, we can plug in all the values in the equation.

*We must remember to use Bedmas when solving the equation.*

Back to the equation: t_n={ar^{n-1}}

a = 21

r = 3

n = 8

= t_n = t_8

= t_8={(21)(3)^{8-1}}

= t_8={(21)(3)^{7}}

= t_8={(21)(2187)}

= t_8={45 927}

The eighth term of the geometric sequence is 45 927.

 

 

 

 

Wonder

Posted on by shannonf2015

The Human Condition

The novel Wonder, written by R. J. Palacio, takes place in Upper Manhattan in New York City. The story shows the life of a young boy named August Pullman, also known as Auggie, who has been homeschooled by his mother due to complicated health issues to a cranio-facial abnormality. When August turns ten, his parents want him to attend school, so he can learn how to navigate the world outside of his town house. The day he attends school, kids and other parents give August stares, pointed fingers, and laugh at his facial deformity. Despite the negativity, he continued to keep his head high by standing up for himself. Knowing everyone will continue to treat and see him differently, he continued to focus on his life with his new friends he encountered at school and his own happiness. This novel shows that the circumstances we face do not determine our happiness, it is our positive outlook and great attitude despite the hardship we face.

Image result for wonder rj palacio

http://i.telegraph.co.uk/multimedia/archive/http://i.telegraph.co.uk/multimedia/archive

 

Week 1 – Arithmetic Sequences

Posted on by shannonf2015

 

This is what I have learned so far in the first week of Pre-Calc 11: 

 

My Arithmetic Sequence:

-27, -20, -13, -6, 1

d = 7

d = common difference (In my arithmetic sequence, 7 is being added each time constantly) 

Formula : t_n={t_1+(n-1)}d

Information that is given:

  1. t_{1} = -27  *t_{1} = first term in a sequence 
  2. n (n is the position of term) = 50
  3. d = 7
  4. t_{n} = t_{50}
  5. t_{50} = ?

Part 1: How to find : t_{50} (term 50 of the sequence) 

Step 1 :

Plug in information that’s given in the formula:

Formula: t_n={t_1+(n-1)}d

= t_{50}={-27+(50-1)(7)}

Step 2 : Solve

  • Caution : We have to use BEDMAS when doing the calculations of the formula.

So it would look like this:

= t_{50}={-27+(49)(7)}

= t_{50}={-27+343}

= t_{50}={316} 

 

How to determine general equation of t_{n} : 

Step 1: Place given information into formula.

Formula: t_n={t_1+(n-1)}d

= t_n={-27+(n-1)(-7)}

= t_n={-27+(-7n+7)}

= t_n={-20+(-7n)}

= t_n={-7n-20} 

 

Part 2: Determine the sum of these first 50 terms. 

Step 1 :

-27 + -20 + -13 + -6 + 1 } a series

Formula to find the sum : S_n=\frac{n}{2}{(t_1+t_n)}

Plug in given information:

  1. t_{1} = -27
  2. n (position of term)=50
  3. t_{50} = 316  <—— we got this answer from finding what is t_{50}
  4. S_{20} = ?

= S_{20}=\frac{50}{2}{(-27+316)}

Step 2: Solve (BEDMAS) 

= S_{20}={25}{(289)}

= S_{20}={7,225}