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If x^2 + y^2 = 70 and 8xy=40 what is the value of (x +y)^2?

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3 Answers

8xy=40, xy=5, y=5/x

x^2  +  (5/x)^2  =  70,  x^4   -70x^2  +25  = 0

(x^2)^2   -70x^2   +25  =0,  use quadratic formula to solve for x^2

x^2=35±20(3)^(1/2),  x =  ±(35±20(3)^(1/2))^(1/2)

Just using the positive values of the (±) signs,

=((35+20(3)^(1/2))^(1/2)  +  5/(35+20(3)^(1/2))^(1/2))^2

=35 + 20(3)^(1/2)  +  10 +  25/(35+20(3)^(1/2)

=35 + 20(3)^(1/2)  +  10 +  (25/(35+20(3)^(1/2))*((25/(35-20(3)^(1/2)/(25/(35-20(3)^(1/2))

=35 + 20(3)^(1/2)  +  10  +35  -  20(3)^(1/2)

=80

The remaining 3 solutions are left.

 

 

 

there are a few approaches you could take here but... from the looks of it the easiest is to recognize 2 things:

  x2 + y2 = 70

 xy = 40/8 = 5

Now multiply out the third equation and substitute the values in:

(x+y)2 = x2 + 2xy + y2

regrouping should make this easy to see from the first 2 equations above, I'll leave the final answer to you:

(x2 + y2) + 2*(xy) = ???

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