Which expression is the simplified form of √k^17
@amistre64 Help?
see atttach
What attach?
is it \(\sqrt{k^{17}}\)?
Yes @satellite73
two goes in to 17 eight times, with a remainder of one, so your answer is \(k^8\sqrt{k}\)
hope the method is clear, it always works that way
What Where did 2 Come from?
square root that is "second root"
but there is no 2 in the equation?
if you had the cube root you would divide the exponent by 3
OH GOTCHA!! THANKS!
ah that is because when you write a square root you do not write the index, but is it is two in other words \(\sqrt{a}=\sqrt[2]{a}\) but we omit the two
Gotcha. Thanks!! and one more?
for other roots you have to write the index cube roots, fourth roots etc yw
go ahead we can do it
√72n^13
same idea as before, but this time we need to recall that \(72=36\times 2\) and therefore \(\sqrt{72}=\sqrt{36\times 2}=\sqrt{36}\times \sqrt{2}=6\sqrt{2}\) you don't need to write the steps, that was my explanation
How did the thirty six turn into the 6???
what is the square root of 36?
and where did n go?
Ok gotcha on teh square root thing.
we are not done yet, i am working one piece at a time
OH ok. :)
first we find that \(\sqrt{72}=6\sqrt{2}\) then, just like last time, we say 2 goes in to 13 six times, with a remainder of 1, so \(\sqrt{n^{13}}=n^6\sqrt{n}\)
K gotcha!
putting them together you get \[\sqrt{72n^{13}}=6n^6\sqrt{2n}\]
THANK YOU SOOOO MUCH YOU EXPLAINED IT PERFECTLY!!!!!
yw
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