Find all ordered pairs of real numbers #(x, y)# such that #x^2y^2 + 2xy^2 + 5x^2 + 3y^2 + 10x + 5 = 0#?
Find all ordered pairs of real numbers (x, y) such that #x^2y^2 + 2xy^2 + 5x^2 + 3y^2 + 10x + 5 = 0# .
Find all ordered pairs of real numbers (x, y) such that
3 Answers
The only real values of x and y are
Explanation:
Given:
Regroup so that the terms containing
Remove the common factor:
Regroup so that the terms containing
Remove the common factor:
Please observe that this is a quadratic in
Therefore, we can write x as a function of y, using the quadratic formula:
This can be simplified a bit:
Please notice that the value under the radical is negative for all values of y except 0.
Therefore, the only real values of x and y are
Explanation:
has a minimum at
NOTE:
The stationary points are determined solving
Also the Hessian or
Explanation:
Given:
#x^2y^2+2xy^2+5x^2+3y^2+10x+5=0#
We can rearrange this as a quadratic in
#(y^2+5)x^2+(2y^2+10)x+(3y^2+5) = 0#
This is in the form:
#ax^2+bx+c = 0#
with:
#{(a = y^2+5), (b=2y^2+10), (c=3y^2+5):}#
This has discriminant
#Delta = b^2-4ac#
#color(white)(Delta) = (2y^2+10)^2-4(y^2+5)(3y^2+5)#
#color(white)(Delta) = (4y^4+40y^2+100)-4(3y^4+20y^2+25)#
#color(white)(Delta) = (4y^4+40y^2+100)-(12y^4+80y^2+100)#
#color(white)(Delta) = -8y^4-40y^2#
So if
If
#0 = 5x^2+10x+5 = 5(x^2+2x+1) = 5(x+1)^2#
Hence