2007 iTest Problems/Problem 18

Problem

Suppose that $x^3+px^2+qx+r$ is a cubic with a double root at $a$ and another root at b, where $a$ and $b$ are real numbers. If $p=-6$ and $q=9$ , what is $r$ ?

Solutions

Solution 1

By Vieta's Formulas, $2a+b = 6$ $2ab + a^2 = 9$ From the first equation, $b = -2a+6$ . Substitute into the second equation to get $-4a^2 + 12a + a^2 = 9$ $-3a^2 + 12a - 9 = 0$ $a^2 - 4a + 3 = 0$ Thus, $a = 3$ or $a = 1$ . If $a = 3$ , then $b = 0$ , so by Vieta's Formulas, $r = 0$ . If $a = 1$ , then $b = 4$ , so by Vieta's Formulas, $r = -4$ . The answer is $\boxed{\textbf{(I)}}$ .

Solution 2

The equation of the polynomial with double root $a$ and single root $b$ is $(x-a)^2 (x-b)$ Expanding the polynomial results in $(x^2 - 2ax + a^2)(x-b)$ $x^3 - (2a+b)x^2 + (2ab+a^2)x - a^2b$ Since $p = -6$ and $q = 9$ , we can write a system of equations. $2a+b = 6$ $2ab + a^2 = 9$ Solving for $a$ results in $a = 3$ or $a = 1$ . If $a = 3$ , then $b = 0$ , so $r = 0$ . If $a = 1$ , then $b = 4$ , so $r = -4$ . Thus, the answer is $\boxed{\textbf{(I)}}$ .

2007 iTest (Problems, Answer Key)
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2007 iTest Problems/Problem 18

Contents

Problem

Solutions

Solution 1

Solution 2

See Also