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

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

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Answer 1 · 2016-10-18T18:11:45

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

Explanation:

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

Solve for y:

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

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

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

Write as the product of two binomials.

$x^{2} + 4 (3 - x^{2}) (3 - x^{2}) = 36 a a a$ $x^2+4(3-x^2)(3-x^2)=36color(white)(aaa)$

$x^{2} + 4 (9 - 6 x^{2} + x^{4}) = 36 a a a$ $x^2+4(9-6x^2+x^4)=36color(white)(aaa)$ Multiply the binomials

$x^{2} + 36 - 24 x^{2} + 4 x^{4} = 36 a a a$ $x^2+36-24x^2+4x^4=36color(white)(aaa)$ Distribute the 4

$4 x^{4} - 23 x^{2} = 0 a a a$ $4x^4-23x^2=0color(white)(aaa)$ Combine like terms

$x^{2} (4 x^{2} - 23) = 0 a a a$ $x^2(4x^2-23)=0color(white)(aaa)$ Factor out an $x^{2}$ $x^2$

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

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

$x = 0$ $x=0$ and $x = \pm \frac{\sqrt{23}}{2} a a a$ $x=+-sqrt(23)/2color(white)(aaa)$ Square root each side.

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

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

Hence, the solutions are, $(1) x = 0, y = 3; (2 and 3) x = \pm \frac{\sqrt{23}}{2}, y = - \frac{11}{4}$ $(1) x=0, y=3; (2 and 3) x=+-sqrt23/2, y=-11/4$ .

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

![desmos.com](useruploads.socratic.org)

Answer 2 · 2016-10-18T18:34:54

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

Explanation:

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

Subtract the first equation from the second:

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

Subtract 33 from both sides:

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

Compute the discriminant:

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

Use the quadratic formula:

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

For $y = 3$ $y = 3$ :

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

$x = 0$ $x = 0$

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

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

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

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

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

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

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How do you find the exact solutions to the system y+x^2=3y+x2=3y+x^2=3 and x^2+4y^2=36x2+4y2=36x^2+4y^2=36?

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