Question

Find the real solutions for `{(x^4-6x^2y^2+y^4=1),(4x^3y+4xy^3=1):}`?

Answer 1

See explanation...

Explanation:

Subtract the second equation from the first to get:

`0 = x^4-4x^3y-6x^2y^2-4xy^3+y^4`

Divide through by `x^2y^2` to get:

`0 = (x/y)^2-4(x/y)-6-4(y/x)+(y/x)^2`

`color(white)(0)=(x/y+y/x)^2 - 4(x/y+y/x) - 8`

`color(white)(0)=(x/y+y/x)^2 - 4(x/y+y/x) + 4 - 12`

`color(white)(0)=((x/y+y/x)-2)^2-(2sqrt(3))^2`

`color(white)(0)=(x/y+y/x-2-3sqrt(2))(x/y+y/x-2+3sqrt(2))`

From the two factors, multiplying each by `(x/y)`, we get two quadratic equations in `(x/y)`...

`(x/y)^2+(-2-3sqrt(2))(x/y)+1 = 0`

`(x/y)^2+(-2+3sqrt(2))(x/y)+1 = 0`

From these we get four values for `(x/y)`:

`(x/y) = 1+(3sqrt(2))/2+-sqrt(18+12sqrt(2))/2`
(both positive)

`(x/y) = 1-(3sqrt(2))/2+-sqrt(18-12sqrt(2))/2`
(both negative)

If `(x/y) = c` then `x = cy` and:

`1 = 4x^3y+4xy^3 = 4c(1+c^2)x^4`

Hence:

`x^4 = 1/(4c(1+c^2))`

In order to have Real solutions, we require `c > 0`

So we require:

`(x/y) = 1+(3sqrt(2))/2+-sqrt(18-12sqrt(2))/2`

Due to the symmetry of the derivation in `x` and `y`, these two values are reciprocals of one another.

So we need only consider one of them then allow `x` and `y` to be swapped in the final solution...

Let `c = (x/y) = 1+(3sqrt(2))/2+sqrt(18-12sqrt(2))/2`

Then:

`x = +-root(4)(1/(4c(1+c^2)))" "` and `" "y = cx`

or:

`y = +-root(4)(1/(4c(1+c^2)))" "` and `" "x = cy`