Two positive distinct numbers a and b each di | vedclass

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(C) Let the number be $x$. According to the problem,the difference between the number and its reciprocal is $1$,so $|x - \frac{1}{x}| = 1$.This gives

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$\sqrt{5}$

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(C) Let the number be $x$. According to the problem,the difference between the number and its reciprocal is $1$,so $|x - \frac{1}{x}| = 1$.This gives two equations: $x - \frac{1}{x} = 1$ and $x - \frac{1}{x} = -1$.For $x - \frac{1}{x} = 1$,we have $x^2 - x - 1 = 0$. Using the quadratic formula,$x = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$. Since $x$ must be positive,$a = \frac{1 + \sqrt{5}}{2}$.For $x - \frac{1}{x} = -1$,we have $x^2 + x - 1 = 0$. Using the quadratic formula,$x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$. Since $x$ must be positive,$b = \frac{\sqrt{5} - 1}{2}$.The sum $a + b = \frac{1 + \sqrt{5}}{2} + \frac{\sqrt{5} - 1}{2} = \frac{1 + \sqrt{5} + \sqrt{5} - 1}{2} = \frac{2\sqrt{5}}{2} = \sqrt{5}$.

Two positive distinct numbers $a$ and $b$ each differ from their reciprocal by $1$. The value of $a + b$ is

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