Question

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`.

Answer 1

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)`

Answer 2

`(-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.

Answer 3

`(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`