Question

A natural number when increased by $84$ equals $160$ times its reciprocal. Find the number.

Answer 1

Let the natural number be $x$.

According to the question, we have

$x+84=160\times \frac{1}{x}$

$\Rightarrow x^2+84x=160$

$\Rightarrow x^2+84x-160=0$

Here a quadratic equation formed :

where $a$ (coefficient of $x^2$) $=1$

$b$ (coefficient of $x$) $=84$

$c$ (constant) $=-160$

D (discriminant) $=b^2-4ac$

$={84}^2-4\left(1\right)(-160)$

$=7696$

$\therefore$ By quadratic formula, we have

$x=\frac{-b\pm \sqrt{D}}{2a}$

$x=\frac{-\left(84\right)\pm \sqrt{7696}}{2}$

$x=\frac{-\left(84\right)-\sqrt{7696}}{2}$ or, $x=\frac{-\left(84\right)+\sqrt{7696}}{2}$

$x=\frac{-\left(84\right)-87.72}{2}$ or, $x=\frac{-\left(84\right)+87.72}{2}$

$\therefore x=-85.86$ or, $x=1.86$