The sides a, b, c of the triangle ABC are the roots of the equation $x^{3} - p x^{2} + q x - r = 0.$
Prove that its area is $\frac{1}{4} [p (4 p q - p^{3} - 8 r)]^{1 / 2}$ .

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Solution

Remember
$(x - α) (x - β) (x - γ) = x^{3} - x^{2} S_{1} + x S_{2} - S_{3}$
Since a, b, c are the roots of
$x^{2} - p x^{2} + q x - r = 0$
$s_{1} = a + b + c = p = 2 s$ or $s = p / 2$
$s_{2} = \sum a b = q, s_{3} = a b c = r$
$△^{2} = s (s - a) (s - b) (s - c)$

= $\frac{P}{2} (\frac{P}{2} - a) (\frac{P}{2} - b) (\frac{P}{2} - c)$

= $\frac{P}{2} [(\frac{P}{2})^{2} - (\frac{P}{2})^{2} s_{1} + \frac{P}{2} s_{2} - s_{3}]$

= $\frac{P}{2} [\frac{- p^{3} + 4 p q - 8 r}{8}]$

$∴ △ = \frac{1}{4} [p (4 p q - p^{3} - 8 r)]^{1 / 2}$

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