Question 865791
If x^2+y^2=20 and xy=30, what is the value of (2x+2y)^2?



Thank you in advance!



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First factor out 2 and rewrite things a bit to get



{{{(2x+2y)^2 = (2(x+y))^2}}}



{{{(2x+2y)^2 = 2^2*(x+y)^2}}}



{{{(2x+2y)^2 = 4(x+y)^2}}}



{{{(2x+2y)^2 = 4(x+y)^2}}}



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Now expand out {{{(x+y)^2}}} to get {{{(x+y)^2=x^2 + 2xy + y^2}}}



So we have these 3 terms: x^2, 2xy, y^2



We have 2 of them so far. The outer terms x^2 and y^2 and they are from the equation x^2+y^2=20



We just need a 2xy term. Well we have xy=30, so let's multiply both sides by 2 to get 2xy=60. 



That wraps up what we need. Let's put it all together.



From above, we found that {{{(2x+2y)^2 = 4(x+y)^2}}}. Replace {{{(x+y)^2}}} with {{{x^2 + 2xy + y^2}}} (since {{{(x+y)^2=x^2 + 2xy + y^2}}}) to get {{{(2x+2y)^2 = 4(x^2 + 2xy + y^2)}}}



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Now group up the x^2 and y^2 terms



{{{(2x+2y)^2 = 4(x^2 + 2xy + y^2)}}}



{{{(2x+2y)^2 = 4(x^2+y^2+2xy)}}}



{{{(2x+2y)^2 = 4((x^2+y^2)+2xy)}}}



And replace {{{x^2+y^2}}} with 20 (because we're given x^2+y^2=20)



{{{(2x+2y)^2 = 4((20)+2xy)}}}



{{{(2x+2y)^2 = 4(20+2xy)}}}



Then replace 2xy with 60 (since we just found that 2xy=60)



{{{(2x+2y)^2 = 4(20+2xy)}}}



{{{(2x+2y)^2 = 4(20+60)}}}



Now compute to finish up



{{{(2x+2y)^2 = 4(20+60)}}}



{{{(2x+2y)^2 = 4(80)}}}



{{{(2x+2y)^2 = 320}}}