Solving For Sides II

More generally, let $a$ , $b$ , and $c$ be the roots of $f(x) = x^3 + px^2 + qx + r,$ so $f(x) = (x - a)(x - b)(x - c)$ .

Assuming that $a$ , $b$ , and $c$ form the sides of a triangle, by Heron's formula, $K^2 = s(s - a)(s - b)(s - c),$ where $K$ and $s$ are the area and semi-perimeter of the triangle, respectively.

By Vieta's formulas, $s = (a + b + c)/2 = -p/2$ , and $f \left( -\frac{p}{2} \right) = f(s) = (s - a)(s - b)(s - c).$ Hence, $K^2 = s(s - a)(s - b)(s - c) = -\frac{p}{2} f \left( -\frac{p}{2} \right).$

Here, $f(x) = (x - 5)(x - 12)(x - 13) - 20$ , and $p = -(5 + 12 + 13) = -30$ , so $K^2 = 15 [(15 - 5)(15 - 12)(15 - 13) - 20] = 600.$

Follow-up Question: Is there an easy way to tell that the roots of $f(x) = (x - 5)(x - 12)(x - 13) - 20$ do form the sides of a triangle?

Expanding and simplying gives $x^3 - 30x^2 + 281x - 800 = 0$ . Let $f(x) = x^3 - 30x^2 + 281x - 800$ , then its zeros are $r_1, r_2, r_3$ .

Similarly, $f(\sqrt x) = 0$ has zeros $r_1^2, r_2^2, r_3^2$

$x \sqrt{x} - 30x + 281 \sqrt{x} - 800 = 0 \Rightarrow x(x+281)^2 = (30x+800)^2$

$\Rightarrow x^3 - 338x^2 + 30961x - 640000 = 0$

$\Rightarrow r_1^2 + r_2^2+ r_3^2 = 338, r_1^2 r_2^2 + r_1^2 r_3^2 + r_2^2 r_3^2 = 30961$

$\Rightarrow r_1^4 + r_2^4 + r_3^4 = (r_1^2 + r_2^2+ r_3^2)^2 - 2 ( r_1^2 r_2^2 + r_1^2 r_3^2 + r_2^2 r_3^2) = 338^2 - 2 \cdot 30961$

One of the variations of Heron's formula is $A = \frac {1}{4} \sqrt{ (a^2 + b^2+c^2)^2 - 2(a^4 + b^4 + c^4)}$

$\Rightarrow A^2 = \frac {1}{16} \cdot \left ( 338^2 - 2 (338^2 - 2 \cdot 30961) \right ) = \boxed{600}$

Let $f(x) = (x-5)(x-12)(x-13)-20$ . Then Heron's formula says that $A^2 = s(s-r_1)(s-r_2)(s-r_3)$ where $s = \frac12(r_1+r_2+r_3).$ . By Vieta, $s = 15$ , and $(s-r_1)(s-r_2)(s-r_3) = f(s) = f(15)$ . So in both problems, the answer is $15 f(15)$ . In this problem it equals $\fbox{600}$ .

Can use graphing calculator-( I have TI-83 +)- and find all three roots. Heron's formula gives us the result. We draw the given curve and y=0 to find intersection through calc function.