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> \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> | + | <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. | + | ~ pi_is_3.14592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067 |
== Solution == | == Solution == | ||
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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} | + | <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=\ | + | 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 21:10, 19 September 2023
Problem
Let and
be the roots of the equation
. Suppose that
and
are the roots of the equation
. What is
?
Video Solution
https://youtu.be/3dfbWzOfJAI?t=457
~ pi_is_3.14592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067
Solution
In a quadratic equation of the form , the product of the roots is
(Vieta's Formulas).
Using this property, we have that and
.
- Notice the fact that we never actually found the roots.
Solution 2
Assume without loss of generality that . We can factor the equation
into
. Therefore,
and
. Using these values, we find
and
. By Vieta's formulas,
is the product of the roots of
, which are
and
. Therefore,
.
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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