2006 AMC 10B Problems/Problem 14

Problem

Let $a$ and $b$ be the roots of the equation $x^2-mx+2=0$ . Suppose that $a+\frac1b$ and $b+\frac1a$ are the roots of the equation $x^2-px+q=0$ . What is $q$ ?

$\mathrm{(A) \ } \frac{5}{2}\qquad \mathrm{(B) \ } \frac{7}{2}\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } \frac{9}{2}\qquad \mathrm{(E) \ } 8$

Video Solution

https://youtu.be/3dfbWzOfJAI?t=457

~ pi_is_3.14

Solution

In a quadratic equation of the form $x^2 + bx + c = 0$ , the product of the roots is $c$ Vieta's Formulas.

Using this property, we have that $ab=2$ and

$q = (a+\frac{1}{b})\cdot(b+\frac{1}{a}) = \frac{ab+1}{b} \cdot \frac{ab+1}{a} = \frac{(ab+1)^2}{ab} = \frac{(2+1)^2}{2} = \frac{9}{2} \Rightarrow D$

Notice the fact that we never actually found the roots.

Solution 2

Assume without loss of generality that $m=3$ . We can factor the equation $x^2-3x+2=0$ into $(x-1)(x-2)=0$ . Therefore, $a=1$ and $b=2$ . Using these values, we find $a+\frac1b=\frac32$ and $b+\frac1a=3$ . By Vieta's formulas, $q$ is the product of the roots of $x^2-px+q=0$ , which are $a+\frac1b$ and $b+\frac1a$ . Therefore, $q=\left(a+\frac1b\right)\left(b+\frac1a\right)=\frac32\cdot3=\frac92\Rightarrow D$

2006 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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Vieta's formulas

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2006 AMC 10B Problems/Problem 14

Contents

Problem

Video Solution

Solution

Solution 2

See Also