MHB Can 20(s - t) Be an Integer in a Quartic Equation with No Real Roots?

[anemone]

Hi MHB,

I don't know how to solve the problem below because I found what we are asked to prove is a bit confusing to me and hence I don't know how to formulate a credible way to solve the problem correctly.

Problem:

Let $s\ne t$ be real numbers. The equation $(x^2+20sx+10t)(x^2+20tx+10s)=0$ has no real roots.

State with reason, if $20(s-t)$ is or isn't an integer.

But one can tell if the quartic equation (which is given in factored form, a product of two quadratic functions) has no real roots, then the discriminant of each quadratic function is less than zero. i.e.

$(20s)^2-4(1)(10t)<0$ and $(20t)^2-4(1)(10s)<0$

Subtracting these two inequalities yields

$20(t-s)(10s+10t+1)<0$ or

$20(s-t)(10s+10t+1)>0$ (*)

By expanding the given equation we have

$x^4+20(s+t)x^3+10(s+2st+t)x^2+200(s^2+t^2)x+100ts=0$ (**)

I spent over an hour trying my best to see what relations that (*) and (**) have but I don't know how to go any further...could someone please help me, please

 

[Opalg]

Re: Is 20(s-t) an integer?

Hi MHB,

I don't know how to solve the problem below because I found what we are asked to prove is a bit confusing to me and hence I don't know how to formulate a credible way to solve the problem correctly.

Problem:

Let $x\ne y$ be real numbers. The equation $(x^2+20sx+10t)(x^2+20tx+10s)=0$ has no real roots.

State with reason, if $20(s-t)$ is or isn't an integer.

But one can tell if the quartic equation (which is given in factored form, a product of two quadratic functions) has no real roots, then the discriminant of each quadratic function is less than zero. i.e.

$(20s)^2-4(1)(10t)<0$ and $(20t)^2-4(1)(10s)<0$

Those two inequalities say that $10s^2<t$ and $10t^2<s$. It follows that both $s$ and $t$ are positive. Take the square root of both sides in the second inequality, to get $\sqrt{10}t<\sqrt s.$ Therefore the point $(s,t)$ (and similarly the point $(t,s)$) must lie above the curve $y = 10x^2$ and below the curve $y = \sqrt{\dfrac x{10}}.$ Now use calculus to find the maximum value of $ \sqrt{\dfrac x{10}} - 10x^2.$ I found that the maximum occurs when $x = (16000)^{-1/3}$ and is equal to $\dfrac3{10(16)^{2/3}} \approx 0.04725.$ This is less than $1/20$, so it appears that $20(s-t)$ cannot be an integer.

 

[anemone]

Re: Is 20(s-t) an integer?

Hi Opalg,

Thank you very much for your guidance in this problem. I appreciate your help!:)

 

[Opalg]

Re: Is 20(s-t) an integer?

My previous answer was only partially correct (in fact, it was wrong). It was correct to say that the point $(s,t)$ has to lie in the interior of the region between the curves $y = 10x^2$ and $y = \sqrt{\dfrac x{10}}$. But it was quite wrong to look at the greatest vertical separation between these curves. What we actually need to do is to maximise the difference $|s-t|$. To do that, we need to find where the point $(s,t)$ is as far as possible from the diagonal $y=x$ (the line where $|s-t| = 0$). The way to do that is to find where the tangent to one of the curves is parallel to the diagonal. So we need to differentiate $y=10x^2$ and put the derivative equal to $1$. That gives $20x=1$. So the point $\bigl(\frac1{20},\frac1{40}\bigr)$ lies on the curve, and its coordinates differ by $1/20$. That is the maximum possible difference for $|s-t|$, and it only occurs on the boundary curves. But the inequalities $10s^2<t$ and $10t^2<s$ are strict, so the point $(s,t)$ must lie in the interior of the region between the curves. For any such interior point we must have $|s-t|<1/20$. It follows that $20(s-t)$ cannot be an integer unless it is $0$. But the problem states that $s\ne t$ (actually, it says $x \ne y$ but since there is no $y$ in the problem I think that must be a misprint for $s\ne t$). Therefore $20(s-t)$ cannot be an integer.