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 $m-n$?

Answer 1

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]