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

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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} = \frac{9}{2} \Rightarrow \boxed{D}</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=\frac92\Rightarrow \boxed{D}</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 16:52, 2 November 2021

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

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

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

See Also

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

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