Given x = 1 / √ 7 - √ 5, y = 1 / √ 7 + √ 5, find the value of x ^ 2 + XY + y ^ 2 | Math enthusiast | Mathematical art

Given x = 1 / √ 7 - √ 5, y = 1 / √ 7 + √ 5, find the value of x ^ 2 + XY + y ^ 2

Rationalizing the denominator
x=(√7+√5)/2,y=(√7-√5)/2
Then x + y = √ 7
xy=(7-5)/4=1/2
So the original formula = (X & # 178; + 2XY + Y & # 178;) - XY
=(x+y)²-xy
=(√7)²-1/2
=7-1/2
=13/2

If | x | = 2 / 3, | y | = 3 / 7, and X + y

-3 + 13 = 0
(x-2)^2+(y+3)^2+√z-3=0
So X-2 = 0, y + 3 = 0, Z-3 = 0
The solution is x = 2, y = - 3, z = 3
So (XY) ^ z = (- 6) ^ 3 = - 216

If x + y = 7, X-Y = 3, find the value of XY

x+y=7,x-y=3
The sum of the two is: 2x = 10, x = 5
So y = 7-5 = 2
x×y
=5×2
=10

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