Can You Crack the Polynomial Challenge VII? Prove 4 Distinct Real Solutions!

[anemone]

Let $p,\,q,\,r,\,s,\,t$ be distinct real numbers. Prove that the equation

$(x-p)(x-q)(x-r)(x-s)+(x-p)(x-q)(x-r)(x-t)+(x-p)(x-q)(x-s)(x-t)+(x-p)(x-r)(x-s)(x-t)+(x-q)(x-r)(x-s)(x-t)=0$

has 4 distinct real solutions.

 

[kaliprasad]

Let $p,\,q,\,r,\,s,\,t$ be distinct real numbers. Prove that the equation

$(x-p)(x-q)(x-r)(x-s)+(x-p)(x-q)(x-r)(x-t)+(x-p)(x-q)(x-s)(x-t)+(x-p)(x-r)(x-s)(x-t)+(x-q)(x-r)(x-s)(x-t)=0$

has 4 distinct real solutions.

Spoiler

Let us call the plynominal on LHS = P(x)
Without loss of generality we can take
$p \lt q \lt r \lt s \lt t$
now $P(p) = (p-q)(p-r)(p-s)(p-t) \gt0$
$P(q) = (q-p)(q-r)(q-s)(q-t) \lt0$
$P(r) = (r-p)(r-q)(r-s)(r-t) \gt0$
$P(s) = (s-p)(s-q)(s-r)(s-t) \lt0$
$P(t) = (t-p)(t-q)(t-r)(t-s) \gt0$
the above 4 says that there is a root in each of the regions (p,q) , (q,r),(r,s),(s,t) and hence all 4 roots are different

 

[anemone]

Spoiler

Let us call the plynominal on LHS = P(x)
Without loss of generality we can take
$p \lt q \lt r \lt s \lt t$
now $P(p) = (p-q)(p-r)(p-s)(p-t) \gt0$
$P(q) = (q-p)(q-r)(q-s)(q-t) \lt0$
$P(r) = (r-p)(r-q)(r-s)(r-t) \gt0$
$P(s) = (s-p)(s-q)(s-r)(s-t) \lt0$
$P(t) = (t-p)(t-q)(t-r)(t-s) \gt0$
the above 4 says that there is a root in each of the regions (p,q) , (q,r),(r,s),(s,t) and hence all 4 roots are different

Well done, kaliprasad, and thanks for participating too!

Here is another proof that is the solution of other great mind:

Spoiler

The LHS of the given equation is the derivative of the function $P(x)=(x-p)(x-q)(x-r)(x-s)(x-t)$, which is continuous and has five distinct real roots.