2024 SSMO Speed Round Problems/Problem 3

Problem

The polynomial $x^3 - 15x^2 + 4x + 4$ has distinct real roots $r$ , $s$ , and $t$ . Find the value of $\left|(r^2 + s^2 + t^2)(rst)\right|.$

Solution 1

From Vieta's formulas, we have $r+s+t = 15, rs+rt+st = 4,$ and $rst = -4.$ Now, note that $\begin{align*} \left|\left(r^2+s^2+t^2\right)(rst)\right| &= \left|\left(\left(r+s+t\right)^2-2(rs+rt+st)\right)\right|\\ &=\left((15^2-2\cdot4\right)(-4)\\ &=\left|217\cdot(-4)\right| = \boxed{868}.\end{align*}$

~SMO_Team

Solution 2

By Vieta's, we have that $r^2+s^2+t^2 = (r+s+t)^2 - 2(rs+st+rt) = 15^2 - 2(4) = 217$ and $rst = -4$ . Thus, $|(r^2+s^2+t^2)(rst)| = |217\cdot (-4)| = \boxed{868}$ .

-Vivdax

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2024 SSMO Speed Round Problems/Problem 3

Problem

Solution 1

Solution 2