The graphs of $y=x^4$ and $y=7x^2-10-5x^2+19/2$ intersect at four points with $x$-coordinates $\pm \sqrt{m}$ and $\pm \sqrt{n}$, where $m > n$. What is $m-n$?
We begin by simplifying the equation $y=7x^{2}-10-5x^{2}+\frac{19}{2}$. Combining like terms yields $y=2x^{2}-10+\frac{19}{2}$. We notice that this is a quadratic equation, so we attempt to factor it as $(x+a)(x+b)$ for some real numbers $a$ and $b$. We have \[y=(x+a)(x+b)=x^{2}+(a+b)x+ab.\]Setting the coefficients equal gives $a+b=0$ and the constant term is $ab=-10+\frac92=-\frac12$. Since $m$ and $n$ are the squares of the roots, Vieta's formulas give \[m+n=(a+b)^2-2ab=a^2+b^2,\]where $ab=-\frac12$. By squaring $a+b=0$, we have $a^2+2ab+b^2=0^2=0$. Simplifying gives $a^2+b^2=-2ab$, so $m+n=-2ab=\frac12$. Because $m>n$, we must have $m=\frac14$ and $n=0$, so $m-n=\boxed{\frac14}$.[1]
To find the four points of intersection between the two graphs, we need to set the equations equal to each other and solve for $x$.
Setting $y = x^4$ equal to $y = 7x^2 - 10 - 5x^2 + \frac{19}{2}$, we have
$$x^4 = 7x^2 - 10 - 5x^2 + \frac{19}{2}.$$
Combining like terms, we get
$$x^4 = 2x^2 - \frac{1}{2}.$$
Putting everything on one side, we have
$$x^4 - 2x^2 + \frac{1}{2} = 0.$$
We can rewrite this equation as
$$\left(x^2 - \frac{1}{\sqrt{2}}\right)^2 = 0.$$
Taking the square root of both sides, we have
$$x^2 - \frac{1}{\sqrt{2}} = 0.$$
Solving for $x^2$, we get
$$x^2 = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}.$$
Taking the square root of both sides, we have
$$x = \pm \frac{1}{\sqrt[4]{2}} = \pm \frac{1}{\sqrt{2} \cdot \sqrt[4]{2}} = \pm \frac{1}{\sqrt{2} \cdot 2^{1/4}} = \pm \frac{1}{\sqrt{2} \cdot \sqrt[4]{4}} = \pm \frac{1}{\sqrt[4]{8}} = \pm \frac{\sqrt[4]{8}}{8}.$$
Simplifying, we have
$$x = \pm \frac{\sqrt[4]{8}}{8} = \pm \frac{\sqrt[4]{2^3}}{8} = \pm \frac{\sqrt{2}}{2^2 \cdot 2} = \pm \frac{\sqrt{2}}{2^3}.$$
Therefore, the four $x$-coordinates of the intersection points are $\pm \frac{\sqrt{2}}{8}$ and $\pm \frac{\sqrt{2}}{8}$.
Since $m$ is the larger $x$-coordinate squared, $m = \left(\frac{\sqrt{2}}{8}\right)^2 = \frac{2}{64} = \frac{1}{32}$.
And $n$ is the smaller $x$-coordinate squared, $n = \left(-\frac{\sqrt{2}}{8}\right)^2 = \frac{2}{64} = \frac{1}{32}$.
Therefore, $m - n = \frac{1}{32} - \frac{1}{32} = \boxed{0}$.