Solving Cubic Polynomial: Prove Two Distinct Roots

[anemone]

Let $p,\,q,\,r,\,s,\,t$ be any real numbers and $s\ne 0$.

Prove that the equation $x^3+(p+q+r)x^2+(pq+qr+rp-s^2)x+t=0$ has at least two distinct roots.

 

[kaliprasad]

Let $p,\,q,\,r,\,s,\,t$ be any real numbers and $s\ne 0$.

Prove that the equation $x^3+(p+q+r)x^2+(pq+qr+rp-s^2)x+t=0$ has at least two distinct roots.

Spoiler

let $P(x) = x^3 +(p+q+r)x^2 + (pq+qr+rp-s^2)x + t$
so $\dfrac{dP(x)}{dx} = 3x^2 + 2 (p+q+r) x + (pq+qr+rp-s^2)$
now discriminant
= $4(p+q+r)^2 - 12(pq+qr+rp-s^2)$
= $4(p^2+q^2+r^2 -pq - qr -rp + 3s^2)$
= $2((p-q)^2 + (q-r)^2+(r-p)^2 + 6s^2)$

now as s is not zero the discriminant is not zero or derivative does not have double root so P(x) cannot have 3 same roots hence it has at least 2 distinct roots

 

[anemone]

Spoiler

let $P(x) = x^3 +(p+q+r)x^2 + (pq+qr+rp-s^2)x + t$
so $\dfrac{dP(x)}{dx} = 3x^2 + 2 (p+q+r) x + (pq+qr+rp-s^2)$
now discriminant
= $4(p+q+r)^2 - 12(pq+qr+rp-s^2)$
= $4(p^2+q^2+r^2 -pq - qr -rp + 3s^2)$
= $2((p-q)^2 + (q-r)^2+(r-p)^2 + 6s^2)$

now as s is not zero the discriminant is not zero or derivative does not have double root so P(x) cannot have 3 same roots hence it has at least 2 distinct roots

Very well done, kaliprasad! (Yes)

Thanks for participating!:)