Question

A positive number whose reciprocal equals one less than the number is

• A. $\frac{(1+\sqrt{2})}{2}$
• B. $\frac{(\sqrt{2}-1)}{4}$
• C. $\frac{(1+\sqrt{5})}{2}$
• D. $\frac{(\sqrt{2}+\sqrt{5})}{2}$

Answer 1

The correct option is C $\frac{(1+\sqrt{5})}{2}$

Let $n$ be the number.

Given,
$\Rightarrow \frac{1}{n}=n-1$

$\Rightarrow 1=n^2-n$

$\Rightarrow n^2-n-1=0$

Comparing it with general form of quadratic equation $a{x}^2+bx+c=0$

Here $a=1,b=-1,c=-1$

Roots are given by ,
$n=\frac{-b\pm \sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$

$=\frac{-(-1)\pm \sqrt{{(-1)}^2-(4\cdot 1\cdot (-1\left)\right)}}{2\cdot 1}$

$\Rightarrow \text{roots}=\frac{1\pm \sqrt{5}}{2}$

Since it is given that the number is positive,

$n=\frac{1+\sqrt{5}}{2}$