need help on this problem i got the answer 10^(1/2) and -10^(1/2) but im not sure.. Find the real solution(s) of the equation involving rational exponents. Check your solutions
(x^2 + 17)^2/3 = 9

Question

need help on this problem i got the answer 10^(1/2) and -10^(1/2) but im not sure.. Find the real solution(s) of the equation involving rational exponents. Check your solutions 
(x^2 + 17)^2/3 = 9

amistre64 · Answer

$${(x^2 + 17)^2\over3} = 9 $$
this equation?

amistre64 · Answer

$$(x^2 + 17)^{2/3} = 9 $$or this one?

anonymous · Answer

the 2nd one

anonymous · Answer

9^(3/2)= 27
so  27= x^2+17
10=x^2
thats what i got

amistre64 · Answer

cube both sides to begin with right?

(x^2 +17)^2 = 729  ; sqrt the sides now

x^2 +17 = +- sqrt(729)   ; -17

x^2 = -17 +- sqrt(729) is what im looking at

anonymous · Answer

is there a way of writing that as a fraction because my assignment is marking it wrong when i submit it as 10^(1/2)

amistre64 · Answer

sqrt the sides again to get:

x = $\large \pm\sqrt{-17 \pm \sqrt{729}}$  :)

anonymous · Answer

sqrt of 10

amistre64 · Answer

the only 'real' solution that comes out of the is sqrt(10) = 10^(1/2)

anonymous · Answer

?

amistre64 · Answer

well, +- sqrt(10); which you had already gotten :)

amistre64 · Answer

what are your options for inputing the answer?  is there a sqrt button?

anonymous · Answer

\sqrt{blah} will give it to you