Week 4 – Multiplying and Dividing Radicals

February 25, 2018February 26, 2018 keishan2015

This week in Pre-Calculus 11 we learned how to simplify expressions involving radicals. We learned how to add/subtract and multiply/divide radicals.

Multiplication:

When multiplying radicals all you need to do is multiply the coefficients of the radicals by each other and the radicands by each other. After doing that you check to see if the radicand can be simplified, and simplify if possible.

Example:

First, in this expression you just multiply the coefficients and radicands together.
You then simplify the radicand if possible. Since the radicand is 32, we look for any perfect squares that could be a factor of it. In this case $3\sqrt{32}$ is the same as ( $\sqrt{16}$ )( $\sqrt{2}$ )
If we know that 16 goes into 32 we take the square root of 16 (4) and then multiply it by the coefficient in front of the radicand. 12(4) = 48.

Example:

First, you need to simplify the expression that is in the brackets before multiplying. To do this all the radicands need to be the same.
Both $3\sqrt{24}$ and $\sqrt{54}$ can be simplified. $3\sqrt{24}$ can be simplified to $6\sqrt{6}$ (the perfect square 4 goes into 24 six times). $\sqrt{54}$ can be simplified to $3\sqrt{6}$ (the perfect square 9 goes into 54 six times).
Since all the radicands in the brackets are the same to simplify all we do is add the like terms. We add or subtract the coefficients and keep the radicands the same (6+3-8 = 1).
We then multiply the two radicals together and simplify.

Division:

When dividing radicals, the only rule is that you CAN’T leave a radical in the denominator as well as a negative.

Conjugate: A conjugate is formed by changing the sign in between two numbers. For example, the conjugate of x + y would be x – y.

Example:

First, we need to understand that there is a radical in the denominator.
To get rid of it we multiply both the numerator and the denominator by the conjugate which in this case would be $\sqrt{5}$ + $\sqrt{3}$ which we write as a fraction. For the fraction to be accurate it should be equivalent to 1.
We the multiply the radicals together and add like terms. In the denominator when you multiply the radicals together there will be some like terms that cancel out ie. $+\sqrt{15}$ and $-\sqrt{15}$ .
After we are left with a fraction that has no radical in its denominator. Though the expression may look like it can’t be further simplified it still can be. The coefficients can still be simplified since they are both divisible by the denominator.

Note: if the denominator was a negative number and the coefficients weren’t divisible by it, then we would divide everything by -1 to get rid of the negative sign.

Week 3 – Absolute Value of a Real Number

February 18, 2018February 19, 2018 keishan2015

This week in Pre-calculus 11 I learned about the absolute values of real numbers and how to solve for them. I learned that the absolute value of a real number is defined as the principal square root of the square of a number.

Principle square root: $\sqrt{36}=6$

The symbol used for the absolute value of a number is two vertical lines: | |

When solving for the absolute value of a number you are saying how far it is away from zero. Whenever you are trying to find the absolute value of a negative number the value will always be positive because, you cant have a negative distance.

ex.

|6| = 6 (6 is 6 numbers away from 0)
|-12| = 12 (-12 is 12 numbers away from 0)
|45| = 45 (45 is 45 numbers away from 0)

If there is an equation inside the absolute value symbol you must solve the equation and find the absolute value of the product.

ex.

|1 – 5| = |-4| = 4
|4 + 6| = |10| = 10
|3 – 9(-2)|= |3 + 18| = |21| = 21

If there is a coefficient in front of the absolute value symbol then you have to first solve for the absolute value and then multiply it by the coefficient. You do not distribute the coefficient inside the symbols.

ex.

5 |2 + 4(-3)|
5 |2 – 12|
5 |-10|
5(10)
50

ex.

– |- $\sqrt{64}$ |
– |-8|
-(8)
-8

Week 2 – Geometric Sequences

February 12, 2018 keishan2015

This week in Pre-calculus 11 I learned about geometric sequences and how to use the formulas to isolate variables. In a geometric sequence each term is multiplied by a constant, known as a common ratio.

My geometric sequence:

3, 12, 48, … 49152

from this sequence we know:

$t_1=3$
$t_n=49152$
r = 4

To find r we use the formula, $r=\frac{t_n}{t_{n-1}}$ , so:

$r=\frac{t_n}{t_{n-1}}$
$r=\frac{12}{3}$
r = 4

In this sequence we are trying to find what n is. To do that we would use the general formula used to find $t_n$ :

$t_n=(t_1)(r)^{n-1}$ , we then input all information that we already know.

$t_n=(t_1)(r)^{n-1}$
$49152=(3)(4)^{n-1}$

we then need to divide everything by 3 to make it easier for us to isolate n.

$[49152=(3)(4)^{n-1}]/3$
$16384=4^{n-1}$

To then find n you keep multiplying 4 by different powers until it matches the value of $t_n$ divided by $t_1$ . In this case we would multiply 4 by different powers until we get 16384.

$16384=4^{7}$

If we know tat the power is 7 then that means that we know, 7 = n -1. We now know that n = 8 because 8 – 1 = 7.

Week 1 – My Arthmetic Sequence (Updated)

February 4, 2018February 8, 2018 keishan2015

This week in Pre-Calculus 11 I learned about sequences and series and the differences between them.

My Arthmetic sequence is:

16, 4, -8, -20, -32, …

In this sequence we know,

$t_1=16$
d = -12 (the sequence decreases by 12 every time)
n = 5 (five is the last known term of the sequence)
$t_n=-32$

Pt. 1

To find the last number of a sequence ( $t_n$ ) a general formula ( $t_n=t_1+(n-1)d$ ) can be altered and simplified to suit my sequence:

$t_n=16+(n-1)(-12)$

$t_n=-12n+28$

So, if we were trying to find the 50th term ( $t_{50}$ ) we would just use the above formula and input all of the information that we know: n = 50 / d = -12/ $t_1=16$