Question

How do you solve the system of equations `y= - 2x ^ { 2}` and `13x ^ { 2} + y ^ { 2} = 12`?

Answer 1

`(sqrt3/2,-3/2),(-sqrt3/2,-3/2),(0+2i,8+0i),(0-2i,8+0i)`

Explanation:

We have

`y=-2x^2`
`13x^2+y^2=12`

Let's substitute the first equation into the second:

`13x^2+(-2x^2)^2=12`

`13x^2+4x^4=12`

`4x^4+13x^2-12=0`

I'll set `A=x^2`:

`4A^2+13A-12=0`

and I'll use the pythagorean formula to solve for roots:

`A = (-b \pm sqrt(b^2-4ac)) / (2a)` with `a=4, b=13, c=-12`

`A=(-13 pm sqrt(13^2-4(4)(-12)))/(2(4))=(-13pm sqrt(169+192))/8=(-13pm19)/8=>`

`=>6/8=3/4`
`=>-32/8=-4`

Since `A=x^2=3/4, -4=>x=pmsqrt(3)/2, pm2i`

We can now solve for resulting `y` values. I'll use `y=-2x^2` :

• `x=sqrt3/2=>y=-2(sqrt3/2)^2=-2(3/4)=-3/2`

• `x=-sqrt3/2=> y=-2(-sqrt3/2)^2=-2(3/4)=-3/2`

The other two roots require graphing across several dimensions using complex numbers:

• `x=2i=>y=-2(2i)^2=-2(-4)=8`

• `x=-2i=>y=-2(-2i)^2=-2(-4)=8`