Question

How do you solve the system of equations `2x^4+y^2=10` and `x^2+2y^4=10`?

Answer 1

The solutions are `S={(sqrt2,sqrt2), (sqrt2,-sqrt2), (-sqrt2,sqrt2),(-sqrt2,-sqrt2)}`

Explanation:

Solve this equation by substitution

`{(2x^4+y^2=10),(x^2+2y^4=10):}`

As the equations are symmetric, we can assume that `x=y`

Therefore,

`2x^4+x^2=10`

`2x^4+x^2-10=0`

Therefore,

`x^2=(-1+-sqrt(1+4*2*10))/(2*2)=(-1+-sqrt81)/(4)`

`=(-1+-9)/4`

`x^2=8/4=2` and `x=-10/4=-5/2`

`x=+-sqrt2` and `x=+-sqrt(5/2)i`

Therefore

When `x=sqrt2`

`2+2y^4=10`

`y^4=4`

`y^2=2`

`y=+-sqrt2`

The solutions `in RR` are

`S={(sqrt2,sqrt2), (sqrt2,-sqrt2), (-sqrt2,sqrt2),(-sqrt2,-sqrt2)}`

The solutions are at the intersection of the `2` curves.

graph{(2x^4+y^2-10)(x^2+2y^4-10)=0 [-10, 10, -5, 5]}