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Find the unique pair of real numbers (x,y) satisfying 

(3x^2 + 6x + 5)(2y^2 + 8y + 10) = 4.

 Jul 18, 2021
 #1
avatar+78 
0

No answer for this since it's very easy to solve, but here's a hint: x and y can be negative, and a negative number squared is positive.

 Jul 18, 2021
 #2
avatar+78 
0

Also, try out imaginary numbers too.

PBJcatalinasandwich  Jul 18, 2021
 #3
avatar+115426 
+1

Find the unique pair of real numbers (x,y) satisfying 

(3x^2 + 6x + 5)(2y^2 + 8y + 10) = 4.

 

 

(3x^2 + 6x + 5)*2(y^2 + 4y + 5) = 4.

(3x^2 + 6x + 5)(y^2 + 4y + 5) = 2

 

try

3x^2 + 6x + 5= 2           and       y^2 + 4y + 5=1

3x^2 + 6x + 3= 0           and       y^2 + 4y + 4= 0

3(x^2+2x+1)=0             and         (y+2)^2=0

3(x+1)^2=0                   and         (y+2)^2=0

 

x=-1,   y=-2

 

This is one solution.  I do not know how to show algebraically that it is unique though.  

 

Here is an interactive graph you can play with if you want.  Graphically, the solution is shown to be unique.

https://www.desmos.com/calculator/jhioscwnvo

 Jul 19, 2021

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