The graphs of $y=x^4$ and $y=5x^2-6$ intersect at four points with $x$-coordinates $\pm \sqrt{m}$ and $\pm \sqrt{n}$, where $m > n$. What is $m-n$?
I want to find the values of $m$ and $n$ that satisfy both equations, so I will set them equal to each other and simplify. I get $x^4 = 5x^2-6$, which I can rearrange to get $x^4-5x^2+6=0$. This looks like a quadratic equation in $x^2$, so I can try to factor it. I look for two numbers that multiply to 6 and add to -5, and I find that -2 and -3 work. So I can write $x^4-5x^2+6 = (x^2-2)(x^2-3)$. This means that either $x^2-2=0$ or $x^2-3=0$, so $x^2 = 2$ or $x^2 = 3$. Taking square roots, I get $x = \pm \sqrt{2}$ or $x = \pm \sqrt{3}$. These are the four $x$-coordinates of the intersection points, so I can identify $m$ and $n$ as 3 and 2 respectively. Therefore, $m-n = 3-2 = 1$. 