Question 9727: If P=x+1/x-1 and Q=x-1/x+1, find P^2+Q^2-PQ.
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! First of all, alway remember to put parentheses to group related terms,
as P= (x+1)/(x-1) and Q=(x-1)/(x+1).
Note, P = 1 + [2/(x-1)] and Q = 1/P = 1 + [-2/(x+1)]
So,P-Q = 2[1/(x-1) + 1/(x+1)] = 2[2x/(x^2-1)]
= 4x/(x^2-1)
and PQ = ??(for you)
Since P^2+Q^2-PQ = (P-Q)^2 +PQ (why ?)
We have P^2+Q^2-PQ = [4x/(x^2-1)]^2 + 1
= 1 + 16x^2 /(x^2-1)^2
Of course ,you can use direct substitution and staightford computation
to get the same asnwer. But, the above way it faster and can avoid some
mistakes.
I won't give further explanations for this level of questions.
Kenny
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