1. ## square roots

It's a square root addition/subtraction but I don't know how to place the symbol for the square root i.e [ math ],

so I'll try my best to describe it.

square root of $3x^2$ plus 6 square root of $108x^2$ minus 4 square root of $108x^2$

Ok here is what I did;

I subtracted + 6 square root of $108x^2$ and -4 square root of $108x^2$

Then I squared off $108x^2$ and got 12 square root of $3x^2$ to which I added to the first square root. and I got with 12 square root of $3x^2$

Now the answer on the answer key is 13x square root of 3. I know I am wrong but I don't know where I am wrong.

Does squaring off the $3x^2$ give me 3 and the x on the outside?

2. If you type &#091;math] \sqrt {108x^2 } [/tex] then you will get $\sqrt {108x^2 }$.

3. Hello, Morzilla!

You did nothing wrong . . . just take another step.

Simplify: . $\sqrt{3x^2} + 6\sqrt{108x^2} - 4\sqrt{ 108x^2}$

we have: . $\sqrt{3x^2} + \underbrace{6\sqrt{108x^2} - 4\sqrt{108x^2}}$

. . . . . . . . $= \;\sqrt{3x^2} + 2\sqrt{108x^2}$

. . . . . . . . $= \;\sqrt{3x^2} + 2\sqrt{36\!\cdot3x^2}$

. . . . . . . . $= \;\sqrt{3x^2} + 2\!\cdot\!6\!\cdot\!\sqrt{3x^2}$

. . . . . . . . $= \;1\!\cdot\!\sqrt{3x^2} + 12\!\cdot\!\sqrt{3x^2}$

. . . . . . . . $= \;13\!\cdot\!\sqrt{3x^2}$ . . . .
This is correct (so far)

. . . . . . . . $= \;13\!\cdot\!\sqrt{3}\!\cdot\!\sqrt{x^2}$

. . . . . . . . $= \;13\!\cdot\sqrt{3}\!\cdot\!x$

. . . . . . . . $= \;13x\sqrt{3}$

4. ## A many, greateful thanks...

Originally Posted by Plato
If you type $\sqrt {108x^2 }$ then you will get $\sqrt {108x^2 }$.
Originally Posted by Soroban
Hello, Morzilla!

You did nothing wrong . . . just take another step.

we have: . $\sqrt{3x^2} + \underbrace{6\sqrt{108x^2} - 4\sqrt{108x^2}}$

. . . . . . . . $= \;\sqrt{3x^2} + 2\sqrt{108x^2}$

. . . . . . . . $= \;\sqrt{3x^2} + 2\sqrt{36\!\cdot3x^2}$

. . . . . . . . $= \;\sqrt{3x^2} + 2\!\cdot\!6\!\cdot\!\sqrt{3x^2}$

. . . . . . . . $= \;1\!\cdot\!\sqrt{3x^2} + 12\!\cdot\!\sqrt{3x^2}$

. . . . . . . . $= \;13\!\cdot\!\sqrt{3x^2}$ . . . .
This is correct (so far)

. . . . . . . . $= \;13\!\cdot\!\sqrt{3}\!\cdot\!\sqrt{x^2}$

. . . . . . . . $= \;13\!\cdot\sqrt{3}\!\cdot\!x$

. . . . . . . . $= \;13x\sqrt{3}$

Many thanks plato!!!

Ahhhhh ok, so there is a 1 in front of the square root!! that makes it 13!!! Thank you! that was part of my question as well!! After looking at it from the start again, I was not 100% sure if a 1 was supposed to be in front of $\sqrt {3x^2 }$

Also, is it me or the text edit options are gone?! the ones to change size, color, boldness and others!?

Thanks