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.