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 $q$?

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

Solution

In a quadratic equation in 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, $q=\left(a+\frac1b\right)\left(b+\frac1a\right)=\frac32\cdot3=\frac92\Rightarrow D$

See Also

2006 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS