Quartic Query

The equation being described is $f(x)=(x-p)(x+p)(x-q)(x+q)=x^4-(p^2+q^2)x^2+p^2q^2$ We want to know the locations of the local extrema, so we take the derivative of this function, and set equal to zero: $0=4x^3-2(p^2+q^2)x$ From this equation and the symmetry of the quartic, zero is clearly an extreme. Dividing out an x leaves a simple quadratic with roots of $\pm \sqrt{(p^2+q^2)/2}$

Next, plug in these values to get the actual heights. After simplifying, 0 gives $p^2q^2$ and $\sqrt{(p^2+q^2)/2}$ gives $-(p^2+q^2)^2/4+p^2q^2$ . Set one height equal to the negative of the other (to account for sign), and you’re done! Well, mostly.

We have an equation relating p and q, now we just need to transform it into a ratio between them. If you’re lazy like me, plug it into a grapher, and you’ll see 4 lines through the origin; the slope of the line where p>q>0 is the answer. But we’ve come this far without a calculator, let’s finish without one.

First, make the replacements $P=p^2$ and $Q=q^2$ . Then, rearrange our equality to get $0= P^2-6PQ+Q^2$ Treat P as the variable and Q as a constant. Plug into the quadratic formula, and get $P=3Q+2\sqrt2Q$ .

P/Q is then $3+2\sqrt2$ . P/Q also happens to be $(p/q)^2$ . Since $3+2\sqrt2=2+2\sqrt2+1=(\sqrt2+1)^2$ our answer is $2+1=\boxed3$

We have $f(x)=(x-p)(x+p)(x-q)(x+q)=x^4-(p^2+q^2)x^2+p^2q^2=(x^2-\frac{p^2+q^2}{2})^2+p^2q^2-\frac{(p^2+q^2)^2}{4}$ , with a minimal value of $p^2q^2-\frac{(p^2+q^2)^2}{4}$ when $x=\pm\sqrt{\frac{p^2+q^2}{2}}$ . We want this minimal value to be $-f(0)=-p^2q^2$ . The equation $p^2q^2-\frac{(p^2+q^2)^2}{4}=-p^2q^2$ simplifies to $4p^2q^2=(p^2-q^2)^2$ and $2pq=p^2-q^2$ . Solving this quadratic equation for $p$ gives $p=(1\pm\sqrt{2})q$ . Since it is required that $p>q$ we have $\frac{p}{q}=\sqrt{2}+1$ and the answer is $\boxed{3}$ .

Let $f(x)=(x^2-p^2)(x^2-q^2)$

Distance of horizontal tangents from x-axis = |Value of function at that point|.

$\therefore, f(0)=|f(c)|$ , for which $f'(c)=0$ (c≠0)

The derivative of this function becomes zero at $x=0,±\sqrt\frac {p^2+q^2}{2}$ .

Putting in the values, we get,

$f(0)=p^2q^2$

$f(c)=-(\frac {p^2-q^2}{2})^2$

Equating them both, and putting $\frac {p}{q}$ as $y$ , we get a quadratic equation,

$y^2-2y-1=0$ , whose solution is $y=1±\sqrt{2}$ . (Only the positive solution is taken as both $p$ and $q$ are >0).