Determine the unique pair of real numbers (x,y) satisfying (4x^2 + 6x + 4)(4y^2 - 12y + 25) = 28
(4x^2 + 6x + 4)(4y^2 - 12y + 25) = 28
(2x^2 + 3x + 2) ( 4y^2 - 12y + 25) = 14
WolframAlpha shows the two solutions as
x = -3/4 y = 3/2
I finally figured this one out without relying on "technology"
(4x^2 + 6x + 4) ( 4y^2 - 12y + 25) = 14 factor out a 2
2 (2x^2 + 6x + 4) (4y^2 - 12 + 25) = 28 divide by 2 on each side
(2x^2 + 3x + 2) ( 4y^2 - 12y + 25) = 14
Factor 2, 4 out of each of the first two terms, respectively
(x^2 + (3/2)x + 1) (y^2 - 3y + 25/4) = 7/4
Complete the square on x and y
( x^2 + (3/2)x + 9/16 + 7/16) ( y^2 - 3y + 9/4 + 4) = 7/4
[ ( x + 3/4)^2 + 7/16 ] [ ( y - 3/2)^2 + 4 ] = 7/4 expand
( x + 3/4)^2 * ( y - 3/2)^2 + 4 ( x + 3/4)^2 + (7/16) ( y - 3/2)^2 + 7/4 = 7/4
Subtract 7/4 from each side
( x + 3/4)^2 * ( y - 3/2)^2 + 4 ( x + 3/4)^2 + (7/16) ( y - 3/2)^2 = 0
Note that each term will equal 0 when x = -3/4 and y = 3/2.....!!!!