# Square root and simplest form

• August 7th 2013, 11:06 AM
uperkurk
Square root and simplest form
$\sqrt{\frac{18ab}{16b^4}} = \sqrt{\frac{9a}{8b^3}}$

Is that in simplest form? There is probably something else I can do but I can't spot anything
• August 7th 2013, 11:29 AM
Plato
Re: Square root and simplest form
Quote:

Originally Posted by uperkurk
$\sqrt{\frac{18ab}{16b^4}} = \sqrt{\frac{9a}{8b^3}}$

Is that in simplest form? There is probably something else I can do but I can't spot anything

It really depends upon you mean by simplest form.

You can say $\sqrt{\frac{18ab}{16b^4}} = \sqrt{\frac{9a}{8b^3}}=\frac{3}{2}\sqrt{\frac{a}{2 b}}$ .
• August 7th 2013, 11:36 AM
uperkurk
Re: Square root and simplest form
Quote:

Originally Posted by Plato
It really depends upon you mean by simplest form.

You can say $\sqrt{\frac{18ab}{16b^4}} = \sqrt{\frac{9a}{8b^3}}=\frac{3}{2}\sqrt{\frac{a}{2 b}}$ .

But would that be classed as "simpler" or just equally similar?

Write in simplest radical form: $\sqrt[3]{54y^8}$ and I have no idea how to go about this.

**EDIT** $\sqrt{(6y\times9y)^4}$ is this a start?
• August 7th 2013, 11:54 AM
Plato
Re: Square root and simplest form
Quote:

Originally Posted by uperkurk
But would that be classed as "simpler" or just equally similar?

Well, I did say it depends on who is asking.

Quote:

Originally Posted by uperkurk
Write in simplest radical form: $\sqrt[3]{54y^8}$ and I have no idea how to go about this.

$\sqrt[3]{54y^8}=\sqrt[3]{(7)(2^3)(y^6)(y^2)}}=2y^2\sqrt[3]{7y^2}$
• August 7th 2013, 11:59 AM
uperkurk
Re: Square root and simplest form
Quote:

Originally Posted by Plato
Well, I did say it depends on who is asking.

$\sqrt[3]{54y^8}=\sqrt[3]{(7)(2^3)(y^6)(y^2)}}=2y^2\sqrt[3]{7y^2}$

I was actually asking you, in your opinion, which would you prefer to see?

also did you mean $\sqrt[3]{(6)(3^3)(y^6)(y^2)}$ ?

Also I can see how you worked it out, thanks.
• August 7th 2013, 12:00 PM
earboth
Re: Square root and simplest form
Quote:

Originally Posted by Plato
Well, I did say it depends on who is asking.

$\sqrt[3]{54y^8}=\sqrt[3]{(7)(2^3)(y^6)(y^2)}}=2y^2\sqrt[3]{7y^2}$

hello,

I don't want to pick at and you but...

$7 \cdot 8 \ne 54$
• August 7th 2013, 12:05 PM
earboth
Re: Square root and simplest form
Quote:

Originally Posted by uperkurk
I was actually asking you, in your opinion, which would you prefer to see?

also did you mean $\sqrt[3]{(6)(3^3)(y^6)(y^2)}$ ?

Also I can see how you worked it out, thanks.

... probably you mean: $\sqrt[3]{(2)(3^3)(y^6)(y^2)}$
• August 7th 2013, 12:09 PM
uperkurk
Re: Square root and simplest form
Quote:

Originally Posted by earboth
... probably you mean: $\sqrt[3]{(2)(3^3)(y^6)(y^2)}$

wha?!!! now I'm mega confused... $(2)(9)=18$

$(6)(9)=54$ ?
• August 7th 2013, 12:11 PM
Plato
Re: Square root and simplest form
Quote:

Originally Posted by earboth
hello,

I don't want to pick at and you but...

$7 \cdot 8 \ne 54$

I don't know what I was thinking. Of course it is 56.

But $54=2(3^3)$ so $\sqrt[3]{54}=3\sqrt[3]2$