Question

How do you find the exact solutions to the system `3x=8y^2` and `8y^2-2x^2=16`?

Answer 1

The possible solutions are:

`(x, y) = (1/4(3+sqrt(119)i), +-sqrt(3/32(3+sqrt(119)i)))`

`(x, y) = (1/4(3-sqrt(119)i), +-sqrt(3/32(3-sqrt(119)i)))`

Explanation:

Given:

`{ (3x=8y^2), (8y^2-2x^2=16) :}`

Substitute `3x` for `8y^2` in the second equation to get:

`3x-2x^2=16`

Add `2x^2-3x` to both sides to get:

`0 = 2x^2-3x+16`

Multiply by `8` to find:

`0 = 16x^2-24x+128`

`color(white)(0) = (4x-3)^2+119`

`color(white)(0) = (4x-3)^2-(sqrt(119)i)^2`

`color(white)(0) = ((4x-3)-sqrt(119)i)((4x-3)+sqrt(119)i)`

`color(white)(0) = (4x-3-sqrt(119)i)(4x-3+sqrt(119)i)`

Hence:

`x = 1/4(3+-sqrt(119)i)`

Then:

`8y^2 = 3x = 3*1/4(3+-sqrt(119)i) = 3/4(3+-sqrt(119)i)`

Divide both ends by `8` to find:

`y^2 = 3/32(3+-sqrt(119)i)`

So the possible solutions are:

`(x, y) = (1/4(3+sqrt(119)i), +-sqrt(3/32(3+sqrt(119)i)))`

`(x, y) = (1/4(3-sqrt(119)i), +-sqrt(3/32(3-sqrt(119)i)))`

It is possible to express `sqrt(3/32(3-sqrt(119)i))` in `a+bi` form if you want, but it gets a little messier:

Using https://socratic.org/s/aw38evei, the principal square root of `a+bi` if it is in Q1 is:

`(sqrt((sqrt(a^2+b^2)+a)/2)) + (sqrt((sqrt(a^2+b^2)-a)/2))i`

I will let you plug in `a=9/32` and `b=(3sqrt(119))/32` if you like.