+0  
 
0
505
2
avatar+644 

 

Determine the unique pair of real numbers (x,y) satisfying (4x^2 + 6x + 4)(4y^2 - 12y + 25) = 28

waffles  Nov 14, 2017
 #1
avatar+87604 
+1

 (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

 

 

cool cool cool

CPhill  Nov 14, 2017
 #2
avatar+87604 
+1

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.....!!!!

 

 

 

cool cool cool

CPhill  Nov 17, 2017

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