2022/018) if a,b,c are roots of equation $x^3-24x^2+183x -440=0$ find the area of the triangle whose sides are a,b,c.

 Note I picked the question at https://www.youtube.com/watch?v=iyX-q3D91Ck&t=474s and here provide a diiferent solution

Let $P(x)= x^3-24x^2+183x -440\cdots(1)$

as a,b,c are roots of the equation  P(x) = 0 so we have following 2 observations

$P(x) = (x-a)(x-b)(x-c) \cdots(2)$

and comparing oefficient of $x^2$ we have

$a+b+c = 24\cdots(3)$

Now area of the triangle say A = $\sqrt{s(s-a)(s-c)(s-c}$

where s is semiperimeter  of triangle 

so $s=\frac{a+b+c}{2}= \frac{24}{2}$ putting value of a+b+c from (3)

or $s= 12\cdots(4)$ 

from (1) putting s for x we get $P(s) = (s-a)(s- b)(s-c)$

So A = $\sqrt{s(s-a)(s-c)(s-c)} = \sqrt{sP(s)}$

so using (1) and above

$A^2 = sP(s) = s(s^3 - 24s^2 + 183 s - 440) $

putting the value of s = 12 from (4) we get

$A^2= 12 *(12^3- 24 * 12^2 + 183 * 12  - 440) = 12(12^3 - 2 * 12^3 + 183 * 12 - 440)$

Or $A^2 = 12(- 12^3 + 183 * 12 - 440) = 12(- 1728 + 2196 - 440) = 12 * 28 = 336$

or $A = \sqrt{336} = \sqrt{4^2 * 21} = 4 \sqrt{21}$