SOLUTION: I don't really know where to place this problem, because it comes in the guide of a contest exam, so I don't know what theme are they talking about, but the problem is like this:

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Question 126344: I don't really know where to place this problem, because it comes in the guide of a contest exam, so I don't know what theme are they talking about, but the problem is like this:
If (x+y)^2=17 and xy=3, then x^2+y^2=
a.20
b.17
c.14
d.11
Can you tell me which theme are they talking about and how to solve it... Thank you...
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%2By%29%5E2=17 Start with the first equation


x%5E2%2B2xy%2By%5E2=17 Foil the left side

Notice how the second equation is xy=3 and the 2nd step has an "2xy" term in it. So let's multiply both sides of xy=3 by 2

2%2Axy=2%2A3

2xy=6



So this means we can replace the "2xy" term in x%5E2%2B2xy%2By%5E2=17 with 6


x%5E2%2B6%2By%5E2=17 Replace 2x with 6


x%5E2%2Bcross%286-6%29%2By%5E2=17-6 Subtract 6 from both sides


x%5E2%2By%5E2=11 Combine like terms



Answer:

So our answer is x%5E2%2By%5E2=11 which means that the answer is D)