2006 AMC 10B Problems/Problem 14: Difference between revisions

Revision as of 21:58, 7 September 2011

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

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

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

@@ Line 12: / Line 12: @@
 == See Also ==
-*[[2006 AMC 10B Problems]]
+{{AMC10 box|year=2006|ab=B|num-b=13|num-a=15}}
-*[[2006 AMC 10B Problems/Problem 13|Previous Problem]]
-*[[2006 AMC 10B Problems/Problem 15|Next Problem]]
 *[[Vieta's formulas]]
 [[Category:Introductory Algebra Problems]]

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2006 AMC 10B Problems/Problem 14: Difference between revisions

Revision as of 21:58, 7 September 2011

Problem

Solution

See Also