3nth sqrt (x-9)^3
\[(\sqrt[3]{x-9})^3\] If you cube a cubed root you just remove them both. \[\sqrt[3]{a^3} = \sqrt[3]{a}^3 = a^{\frac{3}{3}} = a^1 = a \]
its whats inside the radical thats confusing me
x - 9 is just x - 9
Or do I misunderstand your question?
well. i have another problem thats like that and i just dont know what to do with (x-anything)^anything
Just expand it and simplify as much as possible. Unless you have an equal there you can't really solve for x.
so you foil within the radical to expand it?
Well if it's something cubed inside a cubed root, there's no need to foil it cause the two cancel eachother out. Otherwise yeah you'd have to distribute (foil) and hope it turns out nicely.
so the answer is just x-9
Yep.
Assuming that it is as I have it written originally.
thank you, can you help me with sqrt 25(x+2)^4
Of course.
Can you quickly rewrite each of those factors as a square?
25 is what squared?
5
And \((x+2)^4 \) is what squared?
that i dont know
Oh. well what if I asked you what \((2^2)^2\) was. Do you know that?
wouldnt that be 16?
Yes, but don't think about it that way ;p
Think about it as \(2^4\)
When you raise a power to a power you multiply the exponents
Have you been taught this?
the teacher i have sucks. so probably have but not the way your explaining it
Hrm. Ok, well so if raising something to a power multiplies exponents, then taking the root of a power divides the exponent.
For example. \[\sqrt{5^4} = 5^2\] \[\sqrt[3]{7^{12}} = 7^4\] etc.
So if you have \[\sqrt{a^4} =\ ?\]
okaie. so would (x+2)^4 would just be x+2 ^ 2. and the answer to your question is a^2
Yes precisely. So for your problem you had \[\sqrt{5^2(x+2)^4} =\ ?\]
well it was 25 which = 5^2.
Right. Same difference, so take the square root of that and you get?
5(x+2)^2
Exactly.
THANKK YOU SO MUCH<!!!!
Very welcome ! =)
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