Difference between revisions of "2006 AMC 10B Problems/Problem 14"

Revision as of 17:48, 4 August 2023

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 $q$ ?

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

Video Solution

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

~ pi_is_3.14592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067

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

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

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, $q=\left(a+\frac1b\right)\left(b+\frac1a\right)=\frac32\cdot3=\boxed{\textbf{(D) }\frac92}$ .

2006 AMC 10B (Problems • Answer Key • Resources)
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

Vieta's formulas

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

Page

Toolbox

Search