Question

How do you find the exact solutions to the system `y+x^2=3` and `x^2+4y^2=36`?

Answer 1

The solutions are `(0,3)` and `(+-sqrt(23)/2, -11/4)`

Explanation:

`y+x^2=3`

Solve for y:

`y=3-x^2`

Substitute `y` into `x^2+4y^2=36`

`x^2+4(3-x^2)^2=36`

Write as the product of two binomials.

`x^2+4(3-x^2)(3-x^2)=36color(white)(aaa)`

`x^2+4(9-6x^2+x^4)=36color(white)(aaa)`Multiply the binomials

`x^2+36-24x^2+4x^4=36color(white)(aaa)`Distribute the 4

`4x^4-23x^2=0color(white)(aaa)`Combine like terms

`x^2(4x^2-23)=0color(white)(aaa)`Factor out an `x^2`

`x^2=0` and `4x^2-23=0color(white)(aaa)`Set each factor equal to zero

`x^2=0` and `4x^2=23`

`x=0` and `x=+-sqrt(23)/2color(white)(aaa)`Square root each side.

Find the corresponding `y` for each `x` using `y=3-x^2`

`y=3-0=3, and, y=3-23/4=-11/4`

Hence, the solutions are, `(1) x=0, y=3; (2 and 3) x=+-sqrt23/2, y=-11/4`.

Note that there are three solutions, which means there are three points of intersection between the parabola `y+x^2=3` and the ellipse `x^2+4y^2=36`. See the graph below.

![desmos.com]()

Answer 2

Three points of intersection `(-sqrt(23)/2, -11/4)`, `(sqrt(23)/2, -11/4)` and `(0, 3)`

Explanation:

Given:
`y + x^2 = 3`
`x^2 + 4y^2 = 36`

Subtract the first equation from the second:

`4y^2 - y = 33`

Subtract 33 from both sides:

`4y^2 - y - 33 = 0`

Compute the discriminant:

`b^2 - 4(a)(c) = (-1)^2 - 4(4)(-33) = 529`

Use the quadratic formula:

`y = (1 + sqrt(529))/8 = 3` and `y = (1 - sqrt(529))/8 = -11/4`

For `y = 3`:

`x^2 = 3 - 3`

`x = 0`

For `y = -11/4`:

`x^2 = 3 + 11/4`

`x^2 = 12/4 + 11/4`

`x^2 = 23/4`

`x = sqrt(23)/2` and `x = -sqrt(23)/2`