- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,874

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?

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?

Last edited: