suppose x^1/2+x^(-1/2)=3and
b=(x^3/2+x^(-3/2)-3)/(x^2+x^(-2)-2) find the value of b

Question

suppose x^1/2+x^(-1/2)=3and
b=(x^3/2+x^(-3/2)-3)/(x^2+x^(-2)-2)  find the value of b

anonymous · Answer

Ok first cube the intital equation (a+b)^3=a^3+b^3+3ab(a+b)
so x^3/2+x^-3/2 +3(3)=27, so x^3/2+x^-3/2=18.
Now square it you get x^1/2+x^-1/2=7.so substitute in b you get 4/16=1/4

anonymous · Answer

but the ans should be 1/3...

anonymous · Answer

$$x^{1/2}+x ^{-1/2}=3$$
$$x = 1/2 (7+3 \sqrt(5))$$
$$b=(1/2 (1/2 (7-3 \sqrt(5)))^3+(1/2 (7-3 \sqrt(5)))^(-3/2)-3)/((1/2 (7-3 \sqrt(5)))^2+1/(1/2 (7-3 \sqrt(5)))^2-2)$$
$$b=0.332$$

anonymous · Answer

err i think my method is correct only check for calculation mistakes. Write it down you'll get it but i think it is 1/4 only.

anonymous · Answer

Prebz what happened to the other value of x? I think it will still be positive only you cannot neglect it.

anonymous · Answer

Anyway i'm pretty sure thats a very bad method to do it actually finding x and all, all that calculation also, There's no need to find x to get the answer.

anonymous · Answer

The answer should be the same.

anonymous · Answer

How do you say that?

anonymous · Answer

Oh sorry i took the wrong thing in numerator you get 18-3/7-2=3
Now i got 3 not 1/3 :/

anonymous · Answer

Nevermind, the answer will not be the same, it was a bad method.

mertsj · Answer

3ab(a+b) is not 9 from the original equation