Question

find the area of the triangle for which it holds that length of sides are roots of polynomial f(x)= x^3+ax^2+bx+c​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{f(x)=x^2+ax^2+bx+c}$

$\underline{\textbf{To find:}}$

$\textsf{The area of the triangle whose sides are roots of f(x)}$

$\underline{\textbf{Solution:}}$

$\mathsf{Let\;\alpha,\beta\;and\;\gamma\;be\;the\;roots\;of\;f(x)}$

$\mathsf{Then,}$

$\text{\boldmath \alpha+\beta+\gamma=-a }$

$\text{\boldmath \alpha\beta+\beta\gamma+\gamma\alpha=b }$

$\text{\boldmath \alpha\beta\gamma=-c }$

$\mathsf{Clearly\;\alpha,\beta,\gamma\;are\;sides\;of\;the\;required\;triangle}$

$\textsf{We apply Heron's formula to find the area of the triangle}$

$\mathsf{Here,\;s=\dfrac{\alpha+\beta+\gamma}{2}=\dfrac{-a}{2}}$

$\mathsf{Also,}$

$\text{\boldmath (s-\alpha)(s-\beta)(s-\gamma) }$

$\text{\boldmath =s^3-(\alpha+\beta+\gamma)s^2+(\alpha\beta+\beta\gamma+\gamma\alpha)s-\alpha\beta\gamma }$

$\mathsf{=s^3-(-a)s^2+(b)s-c}$

$\mathsf{=s^3+a\,s^2+b\,s-c}$

$\mathsf{=\left(\dfrac{-a}{2}\right)^3+a\left(\dfrac{-a}{2}\right)^2+b\left(\dfrac{-a}{2}\right)-c}$

$\mathsf{=\dfrac{-a^3}{8}+a\left(\dfrac{a^2}{4}\right)-\dfrac{ab}{2}-c}$

$\mathsf{=\dfrac{-a^3+2a^3}{8}-\dfrac{ab}{2}-c}$

$\mathsf{=\dfrac{a^3}{8}-\dfrac{ab}{2}-c}$

$\mathsf{=\dfrac{a^3-4ab-8c}{8}}$

$\mathsf{Now,}$

$\underline{\textsf{Area of the required triangle}}$

$\mathsf{=\sqrt{s(s-a)(s-b)(s-c)}}$

$\mathsf{=\sqrt{\dfrac{-a}{2}\left(\dfrac{a^3-4ab-8c}{8}\right)}}$

$\mathsf{=\sqrt{\dfrac{-a^4+4a^2b+8ac}{16}}}$

$\mathsf{=\dfrac{\sqrt{8ac+4a^2b-a^4}}{4}}}$