Question

The graphs of $x=y^4$ and $x+y^2=1-x$ intersect at two points. The distance between these points is of the form $u+v\sqrt2$, where $u$ and $v$ are integers. Find the ordered pair $(u,v)$.

Answer 1

To find the points of intersection of the graphs $x=y^4$ and $x+y^2=1-x$, we can set these two equations equal to each other:

$y^4 + y^2 + x = 1 - x$

Rearranging the terms, we have:

$y^4 + y^2 + x + x - 1 = 0$

$y^4 + y^2 + 2x - 1 = 0$

Now, let's substitute $x$ with $y^4$:

$y^4 + y^2 + 2y^4 - 1 = 0$

Combining like terms, we get:

$3y^4 + y^2 - 1 = 0$

This is now a quadratic equation in terms of $y^2$. Let's solve it:

Using the quadratic formula, we have:

$y^2 = \frac{-1 \pm \sqrt{1 - 4(3)(-1)}}{2(3)}$

Simplifying,

$y^2 = \frac{-1 \pm \sqrt{13}}{6}$

Taking the square root of both sides,

$y = \pm \sqrt{\frac{-1 \pm \sqrt{13}}{6}}$

Therefore, the two points of intersection are $\left(\sqrt{\frac{-1 + \sqrt{13}}{6}}, \sqrt{\frac{-1 + \sqrt{13}}{6}}\right)$ and $\left(-\sqrt{\frac{-1 + \sqrt{13}}{6}}, -\sqrt{\frac{-1 + \sqrt{13}}{6}}\right)$.

To find the distance between these two points, we can use the distance formula:

$\text{Distance} = \sqrt{\left(\sqrt{\frac{-1 + \sqrt{13}}{6}} - (-\sqrt{\frac{-1 + \sqrt{13}}{6}})\right)^2 + \left(\sqrt{\frac{-1 + \sqrt{13}}{6}} - (-\sqrt{\frac{-1 + \sqrt{13}}{6}})\right)^2}$

Simplifying,

$\text{Distance} = \sqrt{4\left(\frac{-1 + \sqrt{13}}{6}\right)}$

$\text{Distance} = 2\sqrt{\frac{-1 + \sqrt{13}}{6}}$

Therefore, the ordered pair is $(2, \sqrt{-1 + \sqrt{13}})$.