Question

Solve the equation $\dfrac{1}{x}+\dfrac{x}{2}=\dfrac{1}{6}$?

Answer 1

To solve the given equation we have to simplify the given equation furthermore. Certain transformations and substitutions should be made to get the equation simplified and to get the value of $x$. If we get the values of $x$ then we have to verify whether we got the exact values or not.

Complete step-by-step solution:
From the question it had been given that,
$\dfrac{1}{x}+\dfrac{x}{2}=\dfrac{1}{6}$
First of all to get rid of the fractions and unknown in the denominator we have to multiply both sides of the equation with $6x$
By multiplying the given equation with $6x$ we get,
$\Rightarrow \dfrac{6x}{x}+\dfrac{6{{x}^{2}}}{2}=\dfrac{6x}{6}$
$\Rightarrow 6+3{{x}^{2}}=x$
$\Rightarrow 3{{x}^{2}}-x+6=0$
As the simplified equation is in the quadratic equation form, we have to look for the solutions for the quadratic equation. We can get the solutions of the quadratic equation by the quadratic formula.
From the basic concept we know that for any quadratic equation in the form of $a{{x}^{2}}+bx+c=0$ the nature of the solutions is verified by determining the value of $\Delta ={{b}^{2}}-4ac$, the discriminant.
If $\Delta >0$ and a perfect square then there are two real rational solutions.
If $\Delta >0$ and is not a perfect square then there are two real irrational solutions.
If $\Delta <0$ then there are two complex conjugate solutions that are unreal.
If $\Delta =0$ then there is one real solution equal to $\dfrac{-b}{2a}$ .
The equation is in the form of $a{{x}^{2}}+bx+c=0$
Now, by comparing coefficients we get,
$\begin{aligned} & a=3 \\\ & b=-1 \\\ & c=6 \\\ \end{aligned}$
The solutions of the quadratic equation is given by the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
We know that ${{b}^{2}}-4ac$ is the discriminant. First we have to calculate the discriminant.
$\Delta ={{b}^{2}}-4ac$
$\Rightarrow \Delta ={{\left( -1 \right)}^{2}}-4\left( 3 \right)\left( 6 \right)$
$\Rightarrow \Delta =1-72$
$\Rightarrow \Delta =-71$
We can clearly observe that $\Delta <0$.
Therefore, as $\Delta <0$ the given equation has no real solutions.
So the base equation has no real solutions.

Note: While answering questions of this type we should carefully perform the calculations and be sure with the concept. The nature of the zeroes of any quadratic equation is given by the value of $\Delta ={{b}^{2}}-4ac$, the discriminant. If $\Delta >0$ and a perfect square then there are two real rational solutions. If $\Delta >0$ and is not a perfect square then there are two real irrational solutions. If $\Delta <0$ then there are two complex conjugate solutions that are unreal. If $\Delta =0$ then there is one real solution equal to $\dfrac{-b}{2a}$ .