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suppose x^1/2+x^(-1… - QuestionCove
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

but the ans should be 1/3...

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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.

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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.

5 years ago
OpenStudy (anonymous):

The answer should be the same.

5 years ago
OpenStudy (anonymous):

How do you say that?

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (mertsj):

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

5 years ago
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