Question

The sum of a number and its reciprocal is $\dfrac{10}{3}$. Find the number.

Answer 1

Hint: Here, first let us assume that the number be x. Then, its reciprocal will be $\dfrac{1}{x}$. We are given that the sum of the number and its reciprocal is $\dfrac{10}{3}$. Hence, we will get the equation, $x+\dfrac{1}{x}=\dfrac{10}{3}$. Next, by solving we will get a quadratic equation. From the equation find the possible values of x.

Complete step-by-step answer:
$3x(x-3)-x+3=0$
 Next, from the last two terms take -1 outside, we get:
$3x(x-3)-1(x-3)=0$
Since, $x-3$ is the common factor, we can take it outside. Therefore, we obtain:
$(x-3)(3x-1)=0$
We know that if $a.b=0$ then either $a=0$ or $b=0$.
So, we can write:
$x-3=0$ or $3x-1=0$
First consider, $x-3=0$.Here, we are given that the sum of a number and its reciprocal is $\dfrac{10}{3}$.
Now, we have to find the number.
First, let the number be x.
We know that the reciprocal of a number is 1 divided by that number. In other words we can say that if we multiply the number with its reciprocal we will get 1.
Therefore, we can say that the reciprocal of the number x, is $\dfrac{1}{x}$.
Here, it is given if we add the number x and its reciprocal $\dfrac{1}{x}$ we will get $\dfrac{10}{3}$. Hence, we will get:
$x+\dfrac{1}{x}=\dfrac{10}{3}$
 Now by taking the LCM we will get:
$\begin{align}
  & \dfrac{x\times x+1}{x}=\dfrac{10}{3} \\
 & \dfrac{{{x}^{2}}+1}{x}=\dfrac{10}{3} \\
\end{align}$
Next, by cross multiplication we get:
$\begin{align}
  & \left( {{x}^{2}}+1 \right)\times 3=10\times x \\
 & {{x}^{2}}\times 3+1\times 3=10x \\
 & 3{{x}^{2}}+3=10x \\
\end{align}$
Now, by taking 10x to the left hand side, 10x becomes -10x. Hence, we obtain:
$3{{x}^{2}}+3-10x=0$
Next, by rearranging the terms we will get,
$3{{x}^{2}}-10x+3=0$
Hence, we can say that the above equation is a quadratic equation. We know that the quadratic equation has two roots. Next we have to find the two roots of the quadratic equation.
We can solve this by factorisation method. That is, by splitting -10x into -x and -9x. Thus, we obtain:
$3{{x}^{2}}-9x-x+3=0$
Next, from the first two terms, we have the common factor 3x. Therefore, we can take it outside. Hence, we get:

Now, by taking -3 to the right side we get 3. Thus, we obtain:
$x=3$
Next, consider $3x-1=0$.
Now, by taking -1 to the right side we get 1. Therefore, we get
$3x=1$
Now, by cross multiplication we get:
$x=\dfrac{1}{3}$
Therefore, we can say that $x=3$ or $x=\dfrac{1}{3}$.
Hence we can say that the number is either 3 or $\dfrac{1}{3}$.

Note: Here, there are two possibilities for x. If $x=3$ then its reciprocal will be $\dfrac{1}{3}$. Similarly, if $x=\dfrac{1}{3}$ then its reciprocal will be 3. In either cases you will get the product as 1.