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

Revision as of 17:29, 14 July 2006

Problem

Let $a$ and $b$ be the roots of the equation $x^2-mx+2=0$ . Suppose that $a+(1/b)$ and $b+(1/a)$ 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:

$ab=2$

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

@@ Line 1: / Line 1: @@
 == Problem ==
+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+(1/b) </math> and <math> b+(1/a) </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>
 == Solution ==
+In a [[quadratic equation]] in the form <math> x^2 + bx + c = 0 </math>, the product of the roots is <math>c</math>
+Using this property:
+<math>ab=2</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 D </math>
 == See Also ==
 *[[2006 AMC 10B Problems]]

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

Revision as of 17:29, 14 July 2006

Problem

Solution

See Also