AMC 12 Mock Exam 3 Problem 14

Problem 14 of 25

The polynomial $x^3-14x^2+kx-64$ has three real roots that form a geometric progression. What is $k$ ?

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Solution

Write the roots as $\frac ar$ , $a$ , $ar$ . Their product is $a^3=64$ , so $a=4$ . Their sum is $4\left(\frac1r+1+r\right)=14$ , so $r+\frac1r=\frac52$ , giving $r=2$ or $\frac12$ ; either way the roots are $2,4,8$ . Then $k=2\cdot4+4\cdot8+2\cdot8=\boxed{56}$ .

Tagged: cubic with roots in geometric progression (Vieta)

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