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The sum of a number and its reciprocal is $\frac{13}{6}$ Find the number.
The sum of a number and its reciprocal is $\frac{13}{6}$ Find the number.
This Question has 1 answers.
Let the number be $x$.
According to the given condition:
$ x + \frac{1}{x} = \frac{13}{6} $
Multiplying both sides by $x$ to eliminate the fraction:
$ x^2 + 1 = \frac{13}{6} x $
Rearranging the equation:
$ 6x^2 - 13x + 6 = 0 $
Solving the quadratic equation $6x^2 - 13x + 6 = 0$ using the quadratic formula:
$ x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(6)(6)}}{2(6)} $
$ x = \frac{13 \pm \sqrt{169 - 144}}{12} $
$ x = \frac{13 \pm \sqrt{25}}{12} $
$ x = \frac{13 \pm 5}{12} $
Solving for $x$:
$ x = \frac{13 + 5}{12} = \frac{18}{12} = \frac{3}{2} $
$ x = \frac{13 - 5}{12} = \frac{8}{12} = \frac{2}{3} $
The two possible values of the number are $ \frac{3}{2} $ and $ \frac{2}{3} $.
According to the given condition:
$ x + \frac{1}{x} = \frac{13}{6} $
Multiplying both sides by $x$ to eliminate the fraction:
$ x^2 + 1 = \frac{13}{6} x $
Rearranging the equation:
$ 6x^2 - 13x + 6 = 0 $
Solving the quadratic equation $6x^2 - 13x + 6 = 0$ using the quadratic formula:
$ x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(6)(6)}}{2(6)} $
$ x = \frac{13 \pm \sqrt{169 - 144}}{12} $
$ x = \frac{13 \pm \sqrt{25}}{12} $
$ x = \frac{13 \pm 5}{12} $
Solving for $x$:
$ x = \frac{13 + 5}{12} = \frac{18}{12} = \frac{3}{2} $
$ x = \frac{13 - 5}{12} = \frac{8}{12} = \frac{2}{3} $
The two possible values of the number are $ \frac{3}{2} $ and $ \frac{2}{3} $.
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