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

Revision as of 14:00, 2 August 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

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

@@ Line 13: / Line 13: @@
 == See Also ==
 *[[2006 AMC 10B Problems]]
+*[[2006 AMC 10B Problems/Problem 13|Previous Problem]]
+*[[2006 AMC 10B Problems/Problem 15|Next Problem]]
 *[[Vieta's formulas]]

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

Revision as of 14:00, 2 August 2006

Problem

Solution

See Also