Question

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

Answer 1

Hint: Consider the number as $x$ and reciprocal of this number $\dfrac{1}{x}$. By doing certain mathematical operations, convert the equation to quadratic and get the value of x as solution and the number.

Complete step-by-step solution -
From the given question we can write as follows:
$x+84=\dfrac{160}{x}$
By doing mathematical operations we get the equation as,
${{x}^{2}}+84x-160=0$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a)
Since the equation is quadratic, so
$\Delta ={{b}^{2}}-4ac$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (b)
$\Delta ={{84}^{2}}-4\left( 1 \right)\left( -160 \right)$
$\Delta =7696$ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (c)
From above $\Delta$ is positive that means the quadratic equation is having real roots.
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . . . . . . . . . . . . . . . . . . . . . . . . (1)
Substituting values from (a), (b), (c) in (1)
$x=\dfrac{-b\pm \sqrt{\Delta }}{2a}$
$x=\dfrac{-84\pm \sqrt{{{84}^{2}}-4(1)\left( -160 \right)}}{2(1)}$
$x=\dfrac{-84\pm \sqrt{7696}}{2(1)}$
$x=-85.863$
$x=1.863$
Therefore the value of $x$ is $x=-85.863$, $x=1.863$.

Note: Writing the quadratic equation from the given question. From (b) the value of $\Delta$ is positive means the quadratic equation has real roots. If $\Delta$ is negative then the quadratic equation has imaginary roots. If $\Delta$ is 0 then the quadratic equation has equal roots. In this type of question we can use the factorization but if we see the value of x are in decimals so, sometimes it becomes difficult to solve by factorization.