help with radicals

- anonymous

help with radicals

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- schrodinger

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- anonymous

|@LeibyStrauss dw:1434330140647:dw|

- anonymous

|dw:1434330297233:dw|

- anonymous

@LegendarySadist

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## More answers

- UsukiDoll

\[\sqrt{588v^3w^4}\]

- anonymous

@UsukiDoll yea

- UsukiDoll

shall we start on the exponents? it's easier

- anonymous

@UsukiDoll ok

- UsukiDoll

let's start at the fact that we are just dealing with square root so anything even will be square rooted quite nicely. let's start at the \[w^4\]
now.. \[\sqrt{w^4}\] can also be written in exponential form, but since we are not dealing with cube root, square root is 2 and it's not written (that's normal)
in the form of |dw:1434330705533:dw|
\[w^{4/2} \] so what is 4/2 for w ?

- anonymous

w^2

- anonymous

@UsukiDoll w^2

- UsukiDoll

yay! now let's deal with \[\sqrt{v^3}\] we can actually split the exponent up using one of the exponent laws \[v^{a}v^{b}\]
for \[v^3 \rightarrow v^2v\] so \[\sqrt{v^2v}\]

- UsukiDoll

so which one stays in the radical? anything not divisible by 2 stays in the radical . What is the square root of v^2?

- UsukiDoll

what is \[\sqrt{v^2}\] focus on that part only... we can't take square root of v

- anonymous

v^2

- anonymous

is it just radv?

- UsukiDoll

waitttttt.... \[\sqrt{v^3} \] you an split this up \[\sqrt{v^2v}\]

- UsukiDoll

you can take the square root of the v^2 but not the v. so while the only v stays in the radical, you can take the square root of v^2

- UsukiDoll

\[\sqrt{588v^2v(}w^2)\] this is what we have so far.

- UsukiDoll

you can split any odd exponent number up in two.. like for example \[\sqrt{y^5} \rightarrow \sqrt{y^4y} \rightarrow y^2\sqrt{y}\]

- anonymous

ohhh ok makes sense

- UsukiDoll

for square roots we want even exponent numbers... for cube roots we want certain exponent numbers

- UsukiDoll

anyway, now can you tell me what \[\sqrt{v^2v} \] is ?

- anonymous

v^2 sq v

- UsukiDoll

\[\sqrt{v^3} \rightarrow \sqrt{v^2}{v} -> ?????\]
\[\sqrt{v^2} \rightarrow v^{\frac{2}{2}}\]

- UsukiDoll

square root of v is fine, but that v^2 isn't... I have rewritten it in exponential form what is 2/2 ?

- anonymous

just 1

- UsukiDoll

yeah you will have just a v

- UsukiDoll

\[\sqrt{588v(}w^2v)\]
v to the first power or just v

- UsukiDoll

now we have this nasty big number.. we need to not only split this in two but we need the largest perfect square number

- UsukiDoll

otherwise you will be taking radicals over and over again and that takes a lot of time

- anonymous

@UsukiDoll ok how do we do that

- UsukiDoll

perfect square numbers are like 9 16 25 4 36

- anonymous

49*12

- UsukiDoll

let me check that

- UsukiDoll

yeah that works... but we may need to take the radical again for 12 ... eh shouldn't be too bad

- UsukiDoll

\[\sqrt{49 \times 12 [v]} (w^2v)\]

- UsukiDoll

what is the square root of 49 ?

- anonymous

7*7 49 12 4*3

- UsukiDoll

one at a time...

- anonymous

7 for 49

- UsukiDoll

\[\sqrt{12 [v]} (7w^2v) \rightarrow \sqrt{(4)(3)[v]} (7vw^2)\]

- UsukiDoll

mhm now we already have 12 = 4 x 3 so what's the square root of 4

- anonymous

2

- UsukiDoll

\[\sqrt{(3)[v]} (7(2)vw^2)\] we are almost done. we just need to combine 7 and 2.

- UsukiDoll

what is 7 x 2 ?

- anonymous

14

- UsukiDoll

\[\sqrt{3v} (14vw^2)\] yup so this is your final answer

- anonymous

ok thanks usuki

- anonymous

wouldnt the 14vw^2 would go outside

- UsukiDoll

:) I'm just rechecking with wolfram alpha , but this time I don't agree with it . the stupid calculator didn't simplify the w^4 in the square root...
it is outside... the square root is just 3v. everything in the () represented outside.

- anonymous

oh ok :)

- anonymous

yay! thanks!

- UsukiDoll

yes what we did manually is correct... wolfram alpha is making me mad on that v^3 part . the exponent is greater than 2 and can be split up. Wolfram says leave it alone (WRONG!) but purplemath says split it up.
http://www.purplemath.com/modules/radicals2.htm
see the example with the 12abc

- UsukiDoll

the c^3 in that example has been split up to c^2 (c) then that c is left in the radical and the square root of c^2 is taken.

- anonymous

ohh

- anonymous

lol im confused now

- UsukiDoll

wolfram can be sketchy at times... some users on here proven that ultimate calculator to be wrong several times. It shows that not everything is reliable... just focus on the purplemath site and what we did. splitting it up was the right thing to do because that's how I remembered it a long time ago.

- anonymous

ok thanks

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