2023 AMC 12B Problems/Problem 14: Difference between revisions

Revision as of 17:31, 15 November 2023

Denote three roots as $r_1 < r_2 < r_3$ . Following from Vieta's formula, $r_1r_2r_3 = -6$ .

Case 1: All roots are negative.

We have the following solution: $\left( -3, -2, -1 \right)$ .

Case 2: One root is negative and two roots are positive.

We have the following solutions: $\left( -3, 1, 2 \right)$ , $\left( -2, 1, 3 \right)$ , $\left( -1, 2, 3 \right)$ , $\left( -1, 1, 6 \right)$ .

Putting all cases together, the total number of solutions is \boxed{\textbf{(A) 5}}.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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+==Solution==
+Denote three roots as <math>r_1 < r_2 < r_3</math>.
+Following from Vieta's formula, <math>r_1r_2r_3 = -6</math>.
+Case 1: All roots are negative.
+We have the following solution: <math>\left( -3, -2, -1 \right)</math>.
+Case 2: One root is negative and two roots are positive.
+We have the following solutions: <math>\left( -3, 1, 2 \right)</math>, <math>\left( -2, 1, 3 \right)</math>, <math>\left( -1, 2, 3 \right)</math>, <math>\left( -1, 1, 6 \right)</math>.
+Putting all cases together, the total number of solutions is
+\boxed{\textbf{(A) 5}}.
+~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)