Find the unique pair of real numbers (x,y) satisfying
(3x^2 + 6x + 5)(2y^2 + 8y + 10) = 4.
+80 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.
+118673 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.
