Dare to do

Algebra Level 3

Let x and y be real numbers that satisfy the equation x 4 + x 2 y 2 + y 4 = 50 x^{ 4 }+{ x }^{ 2 }{ y }^{ 2 }+{ y }^{ 4 }=50 Then what is the minimum value of 2 x 2 + x y + 2 y 2 2{ x }^{ 2 }+xy+2y^{ 2 } ?

5 5 5\sqrt { 5 } 5 6 5\sqrt { 6 } no answer 6 6 6\sqrt { 6 }

Porames Vattanaprasan by Porames Vattanaprasan , 21, Thailand

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Abhisek Nanda
Sep 22, 2015

minimum of x^2 + xy + y^2 is when x = -y This can be assumed since from equation no 1, x,y belong to (-2.1,2,1) [not exact or accurate) which means negative real numbers can be the values of x and y

Taking x=-y or y =-x , we get equation 1 in problem as 3 X ^ 4 = 50 => x^ 2 = Root (50/3) adding that in the next equation, we get the value as 2x^2 - x*x + 2x^2 = 3x^2= 3 Root of (50/3) = 5 Root (6)

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...