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

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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>?
 
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>\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 </math>
  
 
== Video Solution ==
 
== Video Solution ==
 
https://youtu.be/3dfbWzOfJAI?t=457
 
https://youtu.be/3dfbWzOfJAI?t=457
  
−
~ pi_is_3.14
+
~ pi_is_3.14592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067
  
 
== Solution ==
 
== Solution ==
Line 14: Line 14:
 
Using this property, we have that <math>ab=2</math> and
 
Using this property, we have that <math>ab=2</math> and
  
−
<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>
+
<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} = \boxed{\textbf{(D) }\frac{9}{2}}</math>.
  
 
*Notice the fact that we never actually found the roots.
 
*Notice the fact that we never actually found the roots.
  
 
== Solution 2 ==
 
== Solution 2 ==
−
Assume [[without loss of generality]] that <math>m=3</math>. We can factor the equation <math>x^2-3x+2=0</math> into <math>(x-1)(x-2)=0</math>. Therefore, <math>a=1</math> and <math>b=2</math>. Using these values, we find <math>a+\frac1b=\frac32</math> and <math>b+\frac1a=3</math>. By [[Vieta's formulas]], <math>q</math> is the product of the roots of <math>x^2-px+q=0</math>, which are <math>a+\frac1b</math> and <math>b+\frac1a</math>. Therefore, <math>q=\left(a+\frac1b\right)\left(b+\frac1a\right)=\frac32\cdot3=\frac92\Rightarrow D</math>
+
Assume [[without loss of generality]] that <math>m=3</math>. We can factor the equation <math>x^2-3x+2=0</math> into <math>(x-1)(x-2)=0</math>. Therefore, <math>a=1</math> and <math>b=2</math>. Using these values, we find <math>a+\frac1b=\frac32</math> and <math>b+\frac1a=3</math>. By [[Vieta's formulas]], <math>q</math> is the product of the roots of <math>x^2-px+q=0</math>, which are <math>a+\frac1b</math> and <math>b+\frac1a</math>. Therefore, <math>q=\left(a+\frac1b\right)\left(b+\frac1a\right)=\frac32\cdot3=\boxed{\textbf{(D) }\frac92}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 20:10, 19 September 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, $q=\left(a+\frac1b\right)\left(b+\frac1a\right)=\frac32\cdot3=\boxed{\textbf{(D) }\frac92}$.

See Also

2006 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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All AMC 10 Problems and Solutions

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