Square root and simplest form

$\displaystyle \sqrt{\frac{18ab}{16b^4}} = \sqrt{\frac{9a}{8b^3}}$

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

Re: Square root and simplest form

Quote:

Originally Posted by

**uperkurk** $\displaystyle \sqrt{\frac{18ab}{16b^4}} = \sqrt{\frac{9a}{8b^3}}$

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

Re: Square root and simplest form

Quote:

Originally Posted by

**Plato** It really depends upon you mean by *simplest form*.

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

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

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

**EDIT** $\displaystyle \sqrt{(6y\times9y)^4}$ is this a start?

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

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

Re: Square root and simplest form

Quote:

Originally Posted by

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

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

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

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

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

Re: Square root and simplest form

Quote:

Originally Posted by

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

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

hello,

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

$\displaystyle 7 \cdot 8 \ne 54$

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 $\displaystyle \sqrt[3]{(6)(3^3)(y^6)(y^2)}$ ?

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

... probably you mean: $\displaystyle \sqrt[3]{(2)(3^3)(y^6)(y^2)}$

Re: Square root and simplest form

Quote:

Originally Posted by

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

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

$\displaystyle (6)(9)=54$ ?

Re: Square root and simplest form

Quote:

Originally Posted by

**earboth** hello,

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

$\displaystyle 7 \cdot 8 \ne 54$

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

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