MHB Solving Cubic Polynomial: Prove Two Distinct Roots

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The discussion centers on proving that the cubic polynomial equation \(x^3+(p+q+r)x^2+(pq+qr+rp-s^2)x+t=0\) has at least two distinct roots given that \(s \neq 0\). Participants engage in mathematical reasoning to establish the conditions under which the polynomial exhibits this property. The conversation highlights the significance of the coefficients and their relationships in determining the nature of the roots. The proof is affirmed by a participant, indicating a successful resolution of the problem. The focus remains on the mathematical proof and its implications for cubic equations.
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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.
 
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anemone said:
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.

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
 
kaliprasad said:
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!:)
 
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