anonymous
  • anonymous
help with radicals
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|@LeibyStrauss dw:1434330140647:dw|
anonymous
  • anonymous
|dw:1434330297233:dw|
anonymous
  • anonymous
@LegendarySadist

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UsukiDoll
  • UsukiDoll
\[\sqrt{588v^3w^4}\]
anonymous
  • anonymous
@UsukiDoll yea
UsukiDoll
  • UsukiDoll
shall we start on the exponents? it's easier
anonymous
  • anonymous
@UsukiDoll ok
UsukiDoll
  • 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
  • anonymous
w^2
anonymous
  • anonymous
@UsukiDoll w^2
UsukiDoll
  • 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
  • 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
  • UsukiDoll
what is \[\sqrt{v^2}\] focus on that part only... we can't take square root of v
anonymous
  • anonymous
v^2
anonymous
  • anonymous
is it just radv?
UsukiDoll
  • UsukiDoll
waitttttt.... \[\sqrt{v^3} \] you an split this up \[\sqrt{v^2v}\]
UsukiDoll
  • 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
  • UsukiDoll
\[\sqrt{588v^2v(}w^2)\] this is what we have so far.
UsukiDoll
  • 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
  • anonymous
ohhh ok makes sense
UsukiDoll
  • UsukiDoll
for square roots we want even exponent numbers... for cube roots we want certain exponent numbers
UsukiDoll
  • UsukiDoll
anyway, now can you tell me what \[\sqrt{v^2v} \] is ?
anonymous
  • anonymous
v^2 sq v
UsukiDoll
  • UsukiDoll
\[\sqrt{v^3} \rightarrow \sqrt{v^2}{v} -> ?????\] \[\sqrt{v^2} \rightarrow v^{\frac{2}{2}}\]
UsukiDoll
  • 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
  • anonymous
just 1
UsukiDoll
  • UsukiDoll
yeah you will have just a v
UsukiDoll
  • UsukiDoll
\[\sqrt{588v(}w^2v)\] v to the first power or just v
UsukiDoll
  • 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
  • UsukiDoll
otherwise you will be taking radicals over and over again and that takes a lot of time
anonymous
  • anonymous
@UsukiDoll ok how do we do that
UsukiDoll
  • UsukiDoll
perfect square numbers are like 9 16 25 4 36
anonymous
  • anonymous
49*12
UsukiDoll
  • UsukiDoll
let me check that
UsukiDoll
  • UsukiDoll
yeah that works... but we may need to take the radical again for 12 ... eh shouldn't be too bad
UsukiDoll
  • UsukiDoll
\[\sqrt{49 \times 12 [v]} (w^2v)\]
UsukiDoll
  • UsukiDoll
what is the square root of 49 ?
anonymous
  • anonymous
7*7 49 12 4*3
UsukiDoll
  • UsukiDoll
one at a time...
anonymous
  • anonymous
7 for 49
UsukiDoll
  • UsukiDoll
\[\sqrt{12 [v]} (7w^2v) \rightarrow \sqrt{(4)(3)[v]} (7vw^2)\]
UsukiDoll
  • UsukiDoll
mhm now we already have 12 = 4 x 3 so what's the square root of 4
anonymous
  • anonymous
2
UsukiDoll
  • UsukiDoll
\[\sqrt{(3)[v]} (7(2)vw^2)\] we are almost done. we just need to combine 7 and 2.
UsukiDoll
  • UsukiDoll
what is 7 x 2 ?
anonymous
  • anonymous
14
UsukiDoll
  • UsukiDoll
\[\sqrt{3v} (14vw^2)\] yup so this is your final answer
anonymous
  • anonymous
ok thanks usuki
anonymous
  • anonymous
wouldnt the 14vw^2 would go outside
UsukiDoll
  • 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
  • anonymous
oh ok :)
anonymous
  • anonymous
yay! thanks!
UsukiDoll
  • 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
  • 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
  • anonymous
ohh
anonymous
  • anonymous
lol im confused now
UsukiDoll
  • 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
  • anonymous
ok thanks

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