If x^2 + y^2 = 70 and 8xy = 40

what is the value of (x + y)^2

If x^2 + y^2 = 70 and 8xy = 40

what is the value of (x + y)^2

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Go straight forward:

(x + y)^2

= (x^2 + y^2) + 2(xy)

= 70 + 10, since 8xy = 40, 2xy = 10

= 80

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:

x^{2} + y^{2} = 70

xy = 40/8 = 5

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

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

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

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

## Comments

80 !!

Thank you.