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#.

3 Answers
May 6, 2017

The only real values of x and y are #(-1,0)#

Explanation:

Given:

#x^2y^2 + 2xy^2 + 5x^2 + 3y^2 + 10x + 5 = 0#

Regroup so that the terms containing #x^2# are adjacent:

#color(blue)(x^2)y^2 + 2xy^2 + 5color(blue)(x^2) + 3y^2 + 10x + 5 = 0#

#color(blue)(x^2)y^2 + 5color(blue)(x^2) + 2xy^2 + 3y^2 + 10x + 5 = 0#

Remove the common factor:

#(y^2 + 5)color(blue)(x^2) + 2xy^2 + 3y^2 + 10x + 5 = 0#

Regroup so that the terms containing #x# are adjacent:

#(y^2 + 5)(x^2) + 2color(blue)(x)y^2 + 3y^2 + 10color(blue)(x) + 5 = 0#

#(y^2 + 5)(x^2) + 2color(blue)(x)y^2 + 10color(blue)(x) + 3y^2 + 5 = 0#

Remove the common factor:

#(y^2 + 5)x^2 + (2y^2 + 10)color(blue)(x) + 3y^2 + 5 = 0#

Please observe that this is a quadratic in #color(blue)(x)# where #color(red)(a=(y^2+5))#, #color(green)(b=2y^2+10)# and #color(orange)(c=3y^2+5)#

#color(red)((y^2 + 5))color(blue)(x^2) + color(green)((2y^2 + 10))color(blue)(x) + color(orange)(3y^2 + 5) = 0#

Therefore, we can write x as a function of y, using the quadratic formula:

#x = (-2y^2-10+-sqrt((2y^2 + 10)^2-4(y^2 + 5)(3y^2 + 5)))/(2(y^2 + 5))#

This can be simplified a bit:

#x = -1+-sqrt(1-(3y^2 + 5)/(y^2 + 5))#

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 #(-1,0)#

May 6, 2017

#(-1,0)#

Explanation:

#f(x,y)=x^2y^2 + 2xy^2 + 5x^2 + 3y^2 + 10x + 5 =0#

has a minimum at #-1,0# and at this point

#f(-1,0)=0# so the only real ordered pair is #(-1,0)#

NOTE:

The stationary points are determined solving

#grad f = 0# giving

#((x,y),(-1 - i sqrt[2],-i sqrt[5]),( -1 - i sqrt[2], i sqrt[5]),( -1 + i sqrt[2], -i sqrt[5]),( -1 +i sqrt[2], i sqrt[5]),(-1,0))#

Also the Hessian or #grad^2f# at #(-1,0)# gives

#H=((10,0),(0,4))# which qualifies that point as a local minimum.

May 6, 2017

#(x, y) = (-1, 0)#

Explanation:

Given:

#x^2y^2+2xy^2+5x^2+3y^2+10x+5=0#

We can rearrange this as a quadratic in #x# as:

#(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# given by the formula:

#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 #Delta >= 0# (as required for real roots) then #y=0#

If #y=0# then the given equation reduces to:

#0 = 5x^2+10x+5 = 5(x^2+2x+1) = 5(x+1)^2#

Hence #x=-1#