can someone help me understand on how to simplify radicals?

- anonymous

can someone help me understand on how to simplify radicals?

- Stacey Warren - Expert brainly.com

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- schrodinger

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- saifoo.khan

Where are u stuck? do u have any examples?

- anonymous

how would I simplify 16/81 all under a square root simble?|dw:1327798829785:dw|

- saifoo.khan

\[\sqrt{\frac{16}{81}}\]That right?

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## More answers

- anonymous

yea!!!

- saifoo.khan

That can we written as,
\[\frac{\sqrt{16}}{\sqrt{81}}\]

- saifoo.khan

Can you solve now?

- anonymous

so then it would be like 4 over 9?

- anonymous

is that simplified?

- anonymous

|dw:1327799041137:dw|

- saifoo.khan

Yes 4/9 is correct. but How u got 4/9 ? u simplified using the square root.. So the cancel will be cancelled now.

- anonymous

oh that makes sense, thank you:) would you be able to help me out with a few more problems?

- saifoo.khan

\[\sqrt{\frac{16}{81}} \to \frac{\sqrt{16}}{\sqrt{81}} \to \frac 49 \]

- saifoo.khan

Welcome.
How many do u have?
i will help u as much as i can.

- saifoo.khan

i dont have much time though.

- anonymous

ok, then ill ask you the ones that are more difficult than others

- saifoo.khan

Feel free to do so.

- anonymous

how would i simplify |dw:1327799236709:dw|

- anonymous

and i also am wondering how I would solve\[2\sqrt{3}\div3\sqrt{2}\]

- saifoo.khan

\[\sqrt{75} \text{ can be written as,} \sqrt{ 25 \times 3} \to 5 \sqrt3\]

- anonymous

ok i got that one thanks:)

- saifoo.khan

So now u cann those.

- saifoo.khan

For the second one,
That is simplified, but indirectly.

- anonymous

one more question after the last one i posted is How do i simplify questions that have like 10 over square root 5|dw:1327799591814:dw|

- saifoo.khan

\[\frac{10}{\sqrt5} \times \frac{\sqrt5}{\sqrt5}\]

- anonymous

so for the one thats simplified i would just write simplfyied?

- saifoo.khan

Idk.. have you guys studied rationalizing?

- anonymous

um... i don't know

- saifoo.khan

\[\frac{2\sqrt3}{3 \sqrt 2}\times \frac{3 \sqrt2}{3 \sqrt2}\]Now u may solve this.

- anonymous

i cross multiply?

- saifoo.khan

Nope, simply multiply.
You ONLY cross multiply when there's a "=" sign in between.

- anonymous

ah, i see

- anonymous

Just a nice couple of things about radicals. If a and b are numbers then
\[\sqrt{ab} = \sqrt{a}\sqrt{b}\]
\[\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}\]
and if n and m are any numbers, then
\[n\sqrt{a} + m\sqrt{a} = (n+m)\sqrt{a}\]

- anonymous

should add
\[b \neq 0\]
for that second one..

- anonymous

can you help me with a problem?

- anonymous

what was the second one?

- anonymous

the second one is the fraction, one must never divide by zero. Also a and b are obviously nonnegative (unless it's an odd radical)

- anonymous

um can you help me with a few more questions?

- anonymous

? what's the problem?

- anonymous

how would i simplify |dw:1327800649423:dw|

- saifoo.khan

Rationalize it,
\[\frac{7}{2 \sqrt2} \times \frac{2 \sqrt2}{2 \sqrt2}\]

- anonymous

why do you multiply it?

- saifoo.khan

Whenever u have a sqrt in bottom part. we always do it that way to simplify it.
The method is known as rationalizing.
in which we multiply and divide the fraction by the denomiator.

- anonymous

ok, so im stupid, so can you help me multiply that?

- saifoo.khan

http://www.youtube.com/watch?v=gY5TvlHg4Vk

- anonymous

ok thanks, thats helpful :)

- saifoo.khan

(:

- saifoo.khan

Bye, c ya.

- anonymous

c ya too

- anonymous

on that last one it was sufficient to just multiply by \[\frac{\sqrt{2}}{\sqrt{2}}\]

- anonymous

which one was that?

- anonymous

is|dw:1327802012909:dw|

- anonymous

Ok, suppose you have a fraction like this
\[\frac{7}{2\sqrt{2}}.\]
You want to rationalise the denominator, which means - make the number on the bottom into something without roots.
Now multiplying this fraction by 1 will not change it. Notice that
\[1 = \frac{\sqrt{2}}{\sqrt{2}}\]
and this is true for anything: something divided by itself equals 1.
Therefore
\[\frac{7}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{7\sqrt{2}}{2(\sqrt{2})^2}\]
Now the squared removes the square root and you are just left with 2: \[(\sqrt{2})^2 = 2\] So we get in the end:
\[\frac{7\sqrt{2}}{4}\]
and this is equal to what you started with, and the change we have made is we have rationalised the denominator.

- anonymous

thank you

- anonymous

That's wrong by the way,
\[\sqrt{1} + \sqrt{1} = 2\sqrt{1}\]
because there is two of them!
and we know that \[\sqrt{1} = 1\]
so
\[\sqrt{1} + \sqrt{1} = 2\sqrt{1} = 2\]

- anonymous

and so the square root of 1 +the square root of 1 is 2?

- anonymous

yes, since the square root of 1 is 1, so it reduces to 1+1!

- anonymous

ah, thank you:) I have now finished my math assignment. You have helped me so much, instead of just giving me the answers you helped me understand it

- anonymous

no problem! Hope it's been a help

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