Trigonometry Forever ✞ : 2012

Sunday, December 9, 2012

How do we simplify radical expressions?

Radical expressions look like this:

√108y³

Now when you see that your just like o.O

But it's not that hard as it's made to look, you just have to know how to solve it.

Just like you'd do when you just have a normal radical like √108 , you first separate the biggest square and number that could multiply to that number under the radical.

So for √108:

√36 * √3

36 is a perfect square and is the biggest one that fits into 108.

So in this case since you still have variables, don't forget to simplify those before you jump to let's say √36 * √3 = 6√3

We have y³, so we'd have to look for the greatest squared number.

Which in this case would be y².

That leads to √y² * √y

So lets put everything together.

√36 * √3 * √y² * √y

Now is when we can simplify and put numbers and variables outside of the radical.

36 and y² are both prefect squares, which would go outside of the radical. 3 and y would stay inside the radical since they're not prefect squares. It would look something like this:

Since √36 = 6

and

√y² = y

↓

The outside would be 6y.

And since the 3 and y are simplified as they're gonna get, they stay the same.

Your result would be:

6y√3y

Note: The examples shown in these lessons on radicals show ALL of the steps in the process. It mayNOT be necessary for you to list EVERY step. As long as you understand the process and can arrive at the correct answer, you are ALL SET!!

Sources:

John Schnatterly: johnschnatterly.blogspot.com
http://www.regentsprep.org/Regents/math/ALGEBRA/AO1/Lsimplify.htm

Sunday, November 18, 2012

How do we solve quadratic inequalities?

Okay so for this question there are two ways to solve.

There's algebraically and graphically. Let's just do algebraically because its the hardest, well at least to me it is.

When you get an inequality, the other side to the inequality is going to be 0.

Now what is most important it when you factor an inequality. Once you do that, that is what determines the answer and the symbol of each answer(s) inequality.

For example:

x² - x - 6 < 0

So now you's factor that and get:

(x - 3) (x + 2) < 0

Now here's the trick to remember. When your symbols are different, the product on the other side has to be negative.

But back to the point, since the symbols are different this means that the answers will has different inequality symbols.

Let's see what our answers would be.

x > -2 and x < 3

You could also test this with 0.

If you were to graph this it would be an "and" solution.

This is going to equal -2 < x < 3

So the area in between the black lines
would be where the inequality
is true.

Sources:

John Schnatterly johnschnatterly.blogspot.com
http://www.regentsprep.org/Regents/math/algtrig/ATE6/Quadinequal.htm
http://www.mathsisfun.com/algebra/inequality-quadratic-solving.html

How do we solve quadratic equations by completing the square?

Well let's discuss this shall we.

When you see a quadratic equation you should or already think of an equation like this:

But instead having the a, b, and c
substituted as numbers.

Now using the "completing the square" trick is as easy as it sounds. Actually exactly as it sounds.

Let's say our equation is x² - 4x + 5

Now to solve this equation this way, you would start off by subtracting 5 from both sides.

x² - 4x + 5

- 5

x² - 4x = -5

So now after you do that, you take the value for b (if you forgot which number that is look at the picture above) and divide by 2.

Once you've done that take that answer and square it. That gives you the new square that is know the value for c.

x² - 4x = -5

(-4 divided by 2 is -2)

(and -2 squared is 4)

(so you would add 4 to each side)

x² - 4x = -5

+4 +4

x² - 4x + 4 = -1

And that's simply how you get the new equation to solve this equation.

But this is the most important step to solving a quadratic equation by completing the square.

It's that simple. Now if you want to keep on and solve the equation, be my guest ! :)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Okay i'll do one. But i'll keep doing the same one !

Now you would factor the equation, right.

x² - 4x + 4 = -1

(both -2 and -2 add to -4 and multiply to 4)

( x - 2 )² = -1

[the factor is squared because it would be multiplied by the same factor]

---------------------------------------------------------------------------------------------------------------------------

SO now you would square root both sides to get rid of the squared factor.

√( x - 2 )² = √-1

(now that leaves x-2 by itself)

x-2 = √-1

And i would say that this is where it gets complicated. Here you subtract 2 to each side.

x-2 = √-1

-2 -2

(now you'd put the number subtracted infront an use the ± symbol)

x = -2 ± √-1

& that'd be your answer !!! Hope this helped.

Sources:

Sunday, October 28, 2012

How do we solve quadratic equations?

What are quadratic equations for that matter?

Let's find out.

So a quadratic equation is an equation where the highest coefficient's variable is a square.

In this example, a is the highest variable
because it's variable is
squared.

There are 2 best ways to solve a quadratic equation:

Factoring the equation - when factoring you have to try to put the equation into two parts that can be multiplied and equal the equation. For example :

x² - 8x + 15

(*first find two numbers that multiply to 15, but add to -8*)

(x - 5) (x - 3)

(*you get -5 and -3, which both multiply to 15 and add to -8*)

So your answers would be 5 and 3.

Using the quadratic formula - you find the coefficients and plug them into the formula:

For finding the coefficients you just need to look at the equation. Most likely the a will be the number before the x², the b will be the number before the x, and the c is the number by itself.

An example would be :

3x² + 4x - 56

a is 3, b is 4, and c is -56

Now to plug it into the quadratic formula, you just substitute the letters for the numbers.

x= -(4) ± √( 4² - 4(3)(-56))

————————

2(3)

Now you'd get x is equal to

-4 ± √ (16 + 672)

-4 ± √ (688)

——————

Now just a hint, when you get to this part of the equation you'll either get a rational or irrational/imaginary answer. But it depends on the discriminant (b² - 4ac).

So that's about it for solving equations.

Sources:

Sunday, October 21, 2012

Why do we flip the inequality symbol when multiplying by a negative number or solving absolute value inequalities?

Why should we change the inequality symbol when multiplying or dividing by a negative or when doing absolute value? Why?

I'll tell you why.

An inequality is a form of equation that has a range of answers from 2 values.

So getting back to the question, when you are solving an inequality you are going to have one of these symbols >, <, ≥, ≤

Here's something you should remember when using inequalities.

While solving the inequality you could possibly end up with a negative on either side of the equation. For example with -3x - 5 > 10

With that problem you dumb it down and go through the easy steps of simplifying.

-3x - 5 ≥ 10

(add 5 to both sides)

-3x ≥ 15

Okay so when you get to this step and you have a negative on either side, you'll divide but you have to change the sign. This is because when you get your answer and you plug it in you want it to work. If you forget to change the sign, then when you answer is plugged back in it'll be wrong.

Let me show you what i mean with the equation up there.

-3x - 5 ≥ 10

(add 5 to both sides)

-3x ≥ 15

(divide both sides by -3)

x ≤ -5

(and that's your answer)

Now the answer came out to be x ≤ -5 , so if you plug this back in as a substitute then the inequality is true.

-3 (-5) - 5 ≥ 10

(multiply first)

15 - 5 ≥ 10

(subtract next)

10 ≥ 10

(which is true, according to Einstein)

But that's only possible because that sign was changed when the problem was divided by a negative number.

Just incase you still don't understand, i'll find another reason to help explain.

When you have an inequality like let's say 8x + 4 ≤ -28 , you're most likely going to end up dividing by a negative. So you'll have to change the symbol after you divide. This is also because changing the symbol is the same as moving the negative to the other side.

8x + 4 ≤ -28

(subtract 4)

8x ≤ -32

(divide both sides by 8)

x ≥ -4

Now i changed the sign, but i changed it because if you look closely and do it yourself you can see that it's practically the same as just trying to get the negative to one side.

Hopefully i helped and explained it well, if not then i dont know what to tell you. I'd probably recommend that you ask you teacher for help. :)

Sources Cited: