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

Oct 18, 2016

The solutions are $\left(0 , 3\right)$ and $\left(\pm \frac{\sqrt{23}}{2} , - \frac{11}{4}\right)$

#### Explanation:

$y + {x}^{2} = 3$

Solve for y:

$y = 3 - {x}^{2}$

Substitute $y$ into ${x}^{2} + 4 {y}^{2} = 36$

${x}^{2} + 4 {\left(3 - {x}^{2}\right)}^{2} = 36$

Write as the product of two binomials.

${x}^{2} + 4 \left(3 - {x}^{2}\right) \left(3 - {x}^{2}\right) = 36 \textcolor{w h i t e}{a a a}$

${x}^{2} + 4 \left(9 - 6 {x}^{2} + {x}^{4}\right) = 36 \textcolor{w h i t e}{a a a}$Multiply the binomials

${x}^{2} + 36 - 24 {x}^{2} + 4 {x}^{4} = 36 \textcolor{w h i t e}{a a a}$Distribute the 4

$4 {x}^{4} - 23 {x}^{2} = 0 \textcolor{w h i t e}{a a a}$Combine like terms

${x}^{2} \left(4 {x}^{2} - 23\right) = 0 \textcolor{w h i t e}{a a a}$Factor out an ${x}^{2}$

${x}^{2} = 0$ and $4 {x}^{2} - 23 = 0 \textcolor{w h i t e}{a a a}$Set each factor equal to zero

${x}^{2} = 0$ and $4 {x}^{2} = 23$

$x = 0$ and $x = \pm \frac{\sqrt{23}}{2} \textcolor{w h i t e}{a a a}$Square root each side.

Find the corresponding $y$ for each $x$ using $y = 3 - {x}^{2}$

$y = 3 - 0 = 3 , \mathmr{and} , y = 3 - \frac{23}{4} = - \frac{11}{4}$

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 ${x}^{2} + 4 {y}^{2} = 36$. See the graph below.

Oct 18, 2016

Three points of intersection $\left(- \frac{\sqrt{23}}{2} , - \frac{11}{4}\right)$, $\left(\frac{\sqrt{23}}{2} , - \frac{11}{4}\right)$ and $\left(0 , 3\right)$

#### Explanation:

Given:
$y + {x}^{2} = 3$
${x}^{2} + 4 {y}^{2} = 36$

Subtract the first equation from the second:

$4 {y}^{2} - y = 33$

Subtract 33 from both sides:

$4 {y}^{2} - y - 33 = 0$

Compute the discriminant:

${b}^{2} - 4 \left(a\right) \left(c\right) = {\left(- 1\right)}^{2} - 4 \left(4\right) \left(- 33\right) = 529$

$y = \frac{1 + \sqrt{529}}{8} = 3$ and $y = \frac{1 - \sqrt{529}}{8} = - \frac{11}{4}$

For $y = 3$:

${x}^{2} = 3 - 3$

$x = 0$

For $y = - \frac{11}{4}$:

${x}^{2} = 3 + \frac{11}{4}$

${x}^{2} = \frac{12}{4} + \frac{11}{4}$

${x}^{2} = \frac{23}{4}$

$x = \frac{\sqrt{23}}{2}$ and $x = - \frac{\sqrt{23}}{2}$