1. ## system of equations

(1) x^4 = y - 1
(2) 3x^2 = y + 1

find all solutions in (x,y)

2. Solve each equation for y:

$\displaystyle [x^4 = y - 1] \Rightarrow [x^4 + 1 = y]$
$\displaystyle [3x^2 = y + 1] \Rightarrow [3x^2 - 1 = y]$

Since both equations are equal to y, you can set the two equations equal:

$\displaystyle x^4 + 1 = 3x^2 - 1$

Move everything to one side:

$\displaystyle x^4 - 3x^2 + 2 = 0$

Factor:

$\displaystyle (x^2 - 2)(x^2 - 1) = 0$

Set each equal to 0:

$\displaystyle (x^2 - 2) = 0$
$\displaystyle (x^2 - 1) = 0$

Solve for x:

$\displaystyle x = \pm\sqrt{2}$
$\displaystyle x = \pm\sqrt{1}$

Make it look a little nicer:

$\displaystyle x = -\sqrt{2}, \sqrt{2}, -\sqrt{1}, \sqrt{1}$

And finally, you can substitute these back in to get the values for y if you really need them .

3. Your answer checks with this graph, x^4 - 3x^2 + 2 = 0, thanks.