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Let x and y be real numbers such that 2(x^2 + y^2) = x + y + 1.  Find the maximum value of x - y.

 Jun 21, 2021
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x - y = a

x = a + y

 

2(x^2 + y^2) = x + y + 1

2((a + y)^2 + y^2) = (a + y) + y + 1

2(a^2 + 2y^2 + 2ay) = a + 2y + 1

4y^2 + 4ay + 2a^2 = a + 2y + 1

(4)y^2 + (4a-2)y + 2a^2 - a - 1 = 0

 

We're looking for the highest possible a value. 

This equation is only true when the discriminant is non negative. 

b^2 - 4ac >= 0

(4a-2)^2 - 4(4)(2a^2 - a - 1) >= 0

-16a^2 + 20 >=0

-4a^2 + 5 >= 0 

-4a^2 >= -5

a^2 =< 1.25

So the max value of a is sqrt(1.25).

 

 

This could probably be done faster geometrically too. 

 

=^._.^=

 Jun 21, 2021

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