Question

The sum of a number and its reciprocal is \(\frac{13}{6}\). Find the number.

Answer 1

Let the number be \(x\). Its reciprocal is \(\frac{1}{x}\). According to the problem, the sum of the number and its reciprocal is \(\frac{13}{6}\). So, \(x + \frac{1}{x} = \frac{13}{6}\). To solve for \(x\), first combine the terms on the left side: \(\frac{x^2 + 1}{x} = \frac{13}{6}\). Cross-multiply: \(6(x^2 + 1) = 13x\) \(6x^2 + 6 = 13x\) Rearrange into a standard quadratic equation form: \(6x^2 - 13x + 6 = 0\). We can solve this by factorization. We need two numbers whose product is \(6 \times 6 = 36\) and whose sum is -13. These numbers are -9 and -4. \(6x^2 - 9x - 4x + 6 = 0\) \(3x(2x - 3) - 2(2x - 3) = 0\) \((3x - 2)(2x - 3) = 0\). This gives two possible values for x: 1) \(3x - 2 = 0 \Rightarrow 3x = 2 \Rightarrow x = \frac{2}{3}\). 2) \(2x - 3 = 0 \Rightarrow 2x = 3 \Rightarrow x = \frac{3}{2}\). If the number is \(\frac{2}{3}\), its reciprocal is \(\frac{3}{2}\). Their sum is \(\frac{2}{3} + \frac{3}{2} = \frac{4+9}{6} = \frac{13}{6}\). If the number is \(\frac{3}{2}\), its reciprocal is \(\frac{2}{3}\). Their sum is \(\frac{3}{2} + \frac{2}{3} = \frac{9+4}{6} = \frac{13}{6}\). Both values satisfy the condition. So, the number is \(\frac{2}{3}\) or \(\frac{3}{2}\). \[ \boxed{\text{The number is } \frac{2}{3} \text{ or } \frac{3}{2}} \]