Question 1206122
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I'll assume the given equation is {{{a + 1/(3a) = 2}}}
And I'll assume you want to find the value of {{{9a^2 + 1/(a^2)}}}
Please let me know if otherwise.


If my assumptions are correct, then you should be careful to use parenthesis.
Something like a + 1/3a could be misinterpreted as {{{a+(1/3)a}}}
So instead of a + 1/3a, you should type in a + 1/(3a)
Instead of 9a^2+1/a^2, you should type in 9a^2+1/(a^2)


The original equation just involves 'a' without any exponent.
The thing we want to find has 'a' being squared.
A naive approach could be to simply square both sides to see what happens.


{{{a + 1/(3a) = 2}}}


{{{(a + 1/(3a))^2 = 2^2}}}


{{{a^2 + 2*a*(1/(3a)) + (1/(3a))^2 = 4}}} Use formula (p+q)^2 = p^2+2pq+q^2


{{{a^2 + 2/3 + 1/(9a^2) = 4}}} The 'a's cancel out in the middle portion.


{{{a^2 + 1/(9a^2) = 4 - 2/3}}} 


{{{a^2 + 1/(9a^2) = 10/3}}} 


We want to find {{{9a^2 + 1/(a^2)}}}, but we have {{{a^2 + 1/(9a^2)}}} on the left hand side.
The coefficients for the a^2 terms (1 and 9) have been swapped. 
A quick easy fix is to multiply both sides by 9.


{{{a^2 + 1/(9a^2) = 10/3}}} 


{{{9*(a^2 + 1/(9a^2)) = 9*10/3}}} 


{{{9a^2 + 9*(1/(9a^2)) = 90/3}}} 


{{{9a^2 + 1/(a^2) = 30}}} 



Answer: <font color=red size=4>30</font>
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