Question

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

Answer 1

Hint:
Assume, the number is x and its reciprocal is$\dfrac{1}{x}$. Then add them and equate their sum to$\dfrac{{10}}{3}$.Now we will solve this obtained equation to form a quadratic equation. Once we get the quadratic equation, we will factorize it to get the value of x.

Complete step by step solution:
Given the sum of a number and its reciprocal is$\dfrac{{10}}{3}$.
We have to find the number.
Let the number be x then its reciprocal will be$\dfrac{1}{x}$.
Then according to question,
$ \Rightarrow x + \dfrac{1}{x} = \dfrac{{10}}{3}$
On taking LCM, we get-
$ \Rightarrow \dfrac{{{x^2} + 1}}{x} = \dfrac{{10}}{3}$
On cross-multiplication, we get-
$ \Rightarrow 3\left( {{x^2} + 1} \right) = 10x$
On simplifying, we get-
$ \Rightarrow 3{x^2} + 3 = 10x$
On rearranging, we get-
$ \Rightarrow 3{x^2} - 10x + 3 = 0$ -- (i)
Now here the equation we obtained is in quadratic equation so, we can solve the quadratic equation to find the values of x.
On factorization we get-
$ \Rightarrow 3{x^2} - 9x - x + 3 = 0$
On simplifying, we get-
$ \Rightarrow 3x\left( {x - 3} \right) - 1\left( {x - 3} \right) = 0$
On further simplifying, we get-
$ \Rightarrow \left( {3x - 1} \right)\left( {x - 3} \right) = 0$
On equating either multiplication term to$0$, we gat-
$ \Rightarrow x = \dfrac{1}{3}{\text{ or x = 3}}$
If x=$3$ then its reciprocal is $\dfrac{1}{3}$
And if x=$\dfrac{1}{3}$ then its reciprocal is $3$

Answer- The required number is $3$ and its reciprocal is $\dfrac{1}{3}$ or the required number is $\dfrac{1}{3}$and its reciprocal is $3$

Note:
Here we can also solve the quadratic equation using discriminant methods. If the equation in the standard form $a{x^2} + bx + c = 0$ then we find the value of x using formula-
$ \Rightarrow $ $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
So on comparing the standard equation with eq. (i), we get-
$ \Rightarrow $ a=$3$ , b=$ - 10$ and c=$3$
Then putting the values in the formula, we get-
$ \Rightarrow x = \dfrac{{ - \left( { - 10} \right) \pm \sqrt {{{\left( { - 10} \right)}^2} - 4 \times 3 \times 3} }}{{2 \times 3}}$
On simplifying this we get-
$ \Rightarrow x = \dfrac{{10 \pm \sqrt {100 - 36} }}{6}$
On further solving, we get-
$ \Rightarrow x = \dfrac{{10 \pm \sqrt {64} }}{6} = \dfrac{{10 \pm 8}}{6}$
Now we can get two values of x by first taking the plus sign and solving for x. Then taking the negative sign and solving for x-
$ \Rightarrow x = \dfrac{{10 + 8}}{6}{\text{ or }}\dfrac{{10 - 8}}{6}$
On solving, we get-
$ \Rightarrow x = 3{\text{ or }}\dfrac{1}{3}$
Now we can find the reciprocal of the number.