Question

If a proper fraction in its lower form is subtracted from its reciprocal, the difference is $\dfrac{77}{18}$ . What is the proper fraction?

Answer 1

Hint: Proper fractions are those fractions which have a numerator smaller than the denominator and lowest form is that form of fraction, in which fraction can not be further simplified. Here to find the fraction we will try to get a quadratic equation in terms of \[x\] and further we will solve the quadratic equation for the value of \[x\] and further we will assume whether the fraction is proper or improper.

Complete step-by-step answer:
Let, the fraction be \[x\]. then its reciprocal will be equal to $\dfrac{1}{x}$ as reciprocal of a number is a number which gives 1 on multiplication with that number.
Now, it is given that the difference between proper fraction x and its reciprocal that is $\dfrac{1}{x}$ is equal to $\dfrac{77}{18}$.
So, \[x-\dfrac{1}{x}=\dfrac{77}{18}\]
Multiplying both side by 18,
$\begin{align}
  & 18\left( x-\dfrac{1}{x} \right)=\dfrac{77}{18}\cdot 18 \\
 & 18\left( \dfrac{{{x}^{2}}-1}{x} \right)=77 \\
 & 18\left( {{x}^{2}}-1 \right)=77x \\
\end{align}$
Shifting 77\[x\] from right hand side to left hand side, we get
$18{{x}^{2}}-77x-18=0$ …… ( i )
Now, we have a quadratic equation so we have to find the values of \[x\] using the quadratic formula for roots.
To find the roots of quadratic equation of general form $a{{x}^{2}}+bx+c=0$, the quadratic formula is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
Using quadratic formula in equation ( i ), we get
$\begin{align}
  & x=\dfrac{-(-77)\pm \sqrt{{{(-77)}^{2}}-4(18)(-18)}}{2\cdot (18)} \\
 & x=\dfrac{77\pm \sqrt{5929+1296}}{2\cdot (18)} \\
 & x=\dfrac{77\pm \sqrt{7225}}{36} \\
 & x=\dfrac{77\pm 85}{36} \\
 & x=\dfrac{9}{2},-\dfrac{2}{9} \\
\end{align}$
But, \[x\] cannot be equals to $\dfrac{9}{2}$ as it is not a proper fraction in the lowest terms as the numerator that is 9 is greater than the denominator that is equals to 2.
So, $x=-\dfrac{2}{9}$
Hence, the proper fraction is equals to $x=-\dfrac{9}{2}$ and its reciprocal is equals to $\dfrac{1}{x}=-\dfrac{2}{9}$.


Note: Quadratic equation can be solved by using quadratic formula or by factorising it accordingly. Always remember the value obtained from a quadratic formula is proper fraction itself not its reciprocal and always check if one of the values of \[x\] have a numerator greater than denominator then that fraction is not proper fraction.