Question

Radicals?

Answer 1

Go to:
http://www.purplemath.com/modules/radicals.htm

Answer 2

Go to:
http://www.regentsprep.org/regents/math/algebra/ao1/lsimplify.htm

Answer 3

Go to:
http://tutorial.math.lamar.edu/Classes/Alg/Radicals.aspx

Answer 4

Hope that helps!

Answer 5

\[\sqrt{x-2}=4\]
\[\sqrt{x}+1=5\]
\[2\sqrt{x+3}=10\]
\[\sqrt{x-2}+5=6\]
\[\sqrt{x+4}=-3\]

Answer 6

Go to: {calculator}
http://www.mathcelebrity.com/radex.php?num=sqrt%2840x%5E4y%5E8%29&pl=Simplify+Radical+Expression

Answer 7

Thank you!!

Answer 8

Or go to:
http://www.mathwarehouse.com/arithmetic/square-root-calculator.php

Answer 9

Your welcome, medal?

Answer 10

If you don't mind?

Answer 11

Rather than bounce her all over the internet, why not give her some instruction?

Lets start with your first question.

\(\sqrt{x-2}=4\)

What do you think you would do first?

Answer 12

I know I haven't learned about this lesson or unit, so I thought it would be nice to help her? If you don't mind with the following websites?

Answer 13

I don't really know what to even start with honestly.

Answer 14

Thanks!

Answer 15

You can ask @austinL he will sure help you.

Answer 16

Okay, a radical is simply this.

\(\sqrt{x}=x^{1/2}\)

Which means this,

\(\sqrt{x^a}=a^{a/2}\)

So basically, to cancel out a square root, you simply square it.

In our case, we want to do this,

\(\sqrt{x-2}=4\Rightarrow(\sqrt{x-2})^2=4^2\Rightarrow(x-2)=16\)

Do you think you could solve from there?

Answer 17

No that was very confusing for me.

Answer 18

Lol

Answer 19

Step by step then.

To get rid of a square root you need to square both side, both sides to the second power.

\(\sqrt{x-2}=4\)

Okay, so now to get rid of this square root, we need to do like I said above. Take both sides to the second power.

\((\sqrt{x-2})^2=(x-2)\)

Now that we have that, we need to square the other side as well.

\(4^2=16\)

So now we have this as our equation.

\((x-2)=16\)

From there, we need to get the \(x\) by itself. Which means adding 2 to both sides.

\(x-2+2=16+2\)

\(x\cancel{-2}+\cancel{2}=16+2\)

\(x=18\)

Did this make more sense?

Answer 20

Why do we take them to the second power?

Answer 21

It gets rid of the square root so that you can work with the x alone.

Answer 22

So how did you get 16?

Answer 23

\(4^2=4\times4=16\)

Answer 24

I'm still confused. :/

Answer 25

What is it exactly that you are confused about?

Answer 26

I may be able to help

Answer 27

Okay what he was saying you have to square the square root of x+2 to get ride of the square root right but whenever you do something to one side you must must must do it to the other side so once you square both sides you should get. X+2=16 then you can solve for x just like a normal equation

Answer 28

so do i do it by the 2nd power to every equation?

Answer 29

To cancel out any square root, \(\sqrt{~~~~~~}\), yes.

Answer 30

oh.

Answer 31

so im gunna do the next one and tell me if i get it right

Answer 32

Okay :)

Answer 33

Ok i really dont know how to do this. Ive tried to understand it and its honestly just frustrating the pellet out of me lol. Thanks for the help but I just need to come back to it later when I'm less annoyed.

Answer 34

Okay, just remember square both sides when you have just the square root on one side :)

Answer 35

\(\bf \sqrt{x-2}=4\quad \textit{squaring both sides}\\ \quad \\
(\sqrt{x-2})^2=(4)^2\implies \sqrt{(x-2)^2}=16\implies x-2=16\)

Answer 36

@harvardgirl95 confused still?