Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (proofreading)
m
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
Let <math>a</math> and <math>b</math> be the roots of the equation <math> x^2-mx+2=0 </math>. Suppose that <math> a+(1/b) </math> and <math> b+(1/a) </math> are the roots of the equation <math> x^2-px+q=0 </math>. What is <math>q</math>?
+
Let <math>a</math> and <math>b</math> be the roots of the equation <math> x^2-mx+2=0 </math>. Suppose that <math> a+\frac1b </math> and <math> b+\frac1a </math> are the roots of the equation <math> x^2-px+q=0 </math>. What is <math>q</math>?
  
 
<math> \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 </math>
 
<math> \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 </math>
  
 
== Solution ==
 
== Solution ==
−
In a [[quadratic equation]] in the form <math> x^2 + bx + c = 0 </math>, the product of the roots is <math>c</math>.
+
In a [[quadratic equation]] in the form <math> x^2 + bx + c = 0 </math>, the product of the [[root]]s is <math>c</math>.
  
−
Using this property:
+
Using this property, we have that <math>ab=2</math> and
  
−
<math>ab=2</math>.
+
<math> 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 \mathrm{(D) \ } </math>
−
 
 
−
<math> 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 </math>
 
  
 
== See Also ==
 
== See Also ==
 
*[[2006 AMC 10B Problems]]
 
*[[2006 AMC 10B Problems]]

Revision as of 21:02, 30 July 2006

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 $c$.

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

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

See Also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_14&oldid=8886"