Two positive distinct numbers a and b each di | vedclass

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(C) Given that the numbers $a$ and $b$ differ from their reciprocals by $1$,we have the equation $|x - \frac{1}{x}| = 1$.This implies $x - \frac{1}{x}

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$\sqrt{5}$

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(C) Given that the numbers $a$ and $b$ differ from their reciprocals by $1$,we have the equation $|x - \frac{1}{x}| = 1$.This implies $x - \frac{1}{x} = 1$ or $x - \frac{1}{x} = -1$.For $x - \frac{1}{x} = 1$,we get $x^2 - x - 1 = 0$. The roots are $x = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$. Since $a, b > 0$,we take $a = \frac{1 + \sqrt{5}}{2}$.For $x - \frac{1}{x} = -1$,we get $x^2 + x - 1 = 0$. The roots are $x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$. Since $a, b > 0$,we take $b = \frac{\sqrt{5} - 1}{2}$.Thus,$a + b = \frac{1 + \sqrt{5}}{2} + \frac{\sqrt{5} - 1}{2} = \frac{2\sqrt{5}}{2} = \sqrt{5}$.

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

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