# Math Help - What can we do with radican the same?

1. ## What can we do with radican the same?

Hi all,

I'm trying to learn about radicals and roots, and what we can do with them.

I understand the principle is $x^2 * x^4 = x^(2+4) = x^6$

Basically if the base is the same and we are multiplying we just add the powers.

Same powers rules holds true with square roots i.e.

$n\sqrt{x} * n\sqrt{y}= n\sqrt{x * y}$

Now my question is does the power rules still hold true with the radican being the same?

If not what is it?

Can we say ...

$a\sqrt{4x} * b\sqrt{4x} = (a+b)\sqrt{4x}$ ???

2. Originally Posted by Peleus
$n\sqrt{x} * n\sqrt{y}= n\sqrt{x + y}$
NO! $\sqrt{x} \times \sqrt{y} = \sqrt{xy} \neq \sqrt{x+y}$

$\sqrt{x} \times \sqrt{x} = \sqrt{x^2} = x$

3. Just worked out that no that doesn't work, but is there any particular rule for it?

Cheers

Edit - Also realised I made a typo in the original math hence it should be been x*y instead of x+y.

Thanks

4. Originally Posted by Peleus
Hi all,

I'm trying to learn about radicals and roots, and what we can do with them.

I understand the principle is $x^2 * x^4 = x^(2+4) = x^6$

Basically if the base is the same and we are multiplying we just add the powers.

Same powers rules holds true with square roots i.e.

${\color{red}{n\sqrt{x} * n\sqrt{y}= n\sqrt{x + y}}}$

Now my question is does the power rules still hold true with the radican being the same?

If not what is it?

Can we say ...

$a\sqrt{4x} * b\sqrt{4x} = (a+b)\sqrt{4x}$ ???

${\color{red}{n\sqrt{x} * n\sqrt{y}= n\sqrt{x + y}}}$ this is wrong it should be ${n\sqrt{x} + n\sqrt{y}= n\big(\sqrt{x} + \sqrt{y}\big)}$ or ${n\sqrt{x} * n\sqrt{y}= n^{2}\big(\sqrt{x*y}\big)}$
On the other hand, we can not say $a\sqrt{4x} * b\sqrt{4x} = (a+b)\sqrt{4x}$, because this is equal to $a\sqrt{4x} * b\sqrt{4x}=a*\sqrt{4x} * b*\sqrt{4x} = a*b\Big(\sqrt{4x}\Big)^{2}=a*b*4*x$

5. For any real numbers a, b, n, m:
$(ab)^n = a^n b^n$

$\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$

$a^n \times a^m = a^{n+m}$

$\frac{a^n}{a^m} = a^{n-m}$

$\sqrt[n]{a} = a^{\frac{1}{n}}$

6. Originally Posted by Chop Suey
For any real numbers a, b, n, m:
$(ab)^n = a^n b^n$

$\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$

$a^n \times a^m = a^{n+m}$

$\frac{a^n}{a^m} = a^{n-m}$

$\sqrt[n]{a} = a^{\frac{1}{n}}$
Also an important one $(a^{m})^{n}=a^{mn}=(a^{n})^{m}$.

7. Originally Posted by Peleus
Hi all,

I'm trying to learn about radicals and roots, and what we can do with them.

I understand the principle is $x^2 * x^4 = x^(2+4) = x^6$

Basically if the base is the same and we are multiplying we just add the powers.

Same powers rules holds true with square roots i.e.

$n\sqrt{x} * n\sqrt{y}= n\sqrt{x * y}$

Now my question is does the power rules still hold true with the radican being the same?

If not what is it?

Can we say ...

$a\sqrt{4x} * b\sqrt{4x} = (a+b)\sqrt{4x}$ ???

The "ath root" of 4x is $(4x)^{1/a}$ and the "bth root" of 4x is $(4x)^{1/b}$. By the usual formula for multiplying powers, $(4x)^{1/a}(4x)^{1/b}= (4x)^{1/a+ 1/b}= (4x)^{\frac{a+b}{ab}}= ^{ab}\sqrt{(4x)^{a+b}}$