# Solve for $x:2\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)-9\left( x+\dfrac{1}{x} \right)+14=0$.

A. $\dfrac{1}{2},1,2$

B. $\dfrac{1}{2},1,-2$

C. $\dfrac{1}{2},1,4$

D. $\dfrac{1}{2},1,-4$

Answer

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**Hint:**For solving this question you should know about the general mathematical solution of the equations. It means we will calculate the values of x and for this we will take any term as ‘t’ and then solve that ‘t’ and then we will find the solution. We will then again substitute it in the equation and by solving that, we will finally get the answer.

**Complete step-by-step solution:**

According to our question, it is asked of us to solve the equation, $2\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)-9\left( x+\dfrac{1}{x} \right)+14=0$ and get the values of x. as we know that any quadratic equation has two or more than two real or imaginary roots and these are the solutions of this quadratic equation. And if the equation is in any other form, then we will set it in a proper form. And calculate the value of x. So, if we take our question, the equation is:

$2\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)-9\left( x+\dfrac{1}{x} \right)+14=0$

Let $x+\dfrac{1}{x}=t$. So,

On squaring both side we get,

$\begin{align}

& \Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}+2={{t}^{2}} \\

& \Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}={{t}^{2}}-2 \\

\end{align}$

Substituting these values in the equation, we get the following:

$\begin{align}

& 2\left( {{t}^{2}}-2 \right)-9t+14=0 \\

& \Rightarrow 2{{t}^{2}}-4-9t+14=0 \\

& \Rightarrow 2{{t}^{2}}-9t+10=0 \\

& \Rightarrow 2{{t}^{2}}-4t-5t+10=0 \\

& \Rightarrow \left( t-2 \right)\left( 2t-5 \right)=0 \\

\end{align}$

So, the values of t are:

$t=2,t=\dfrac{5}{2}$

So, if we take these values and substitute them back, then we will get:

Case 1, when $t=2$,

$\begin{align}

& \Rightarrow x+\dfrac{1}{x}=2 \\

& \Rightarrow {{\left( x-1 \right)}^{2}}=0 \\

& \Rightarrow x=1 \\

\end{align}$

And case, when $t=\dfrac{5}{2}$

$\begin{align}

& \Rightarrow x+\dfrac{1}{x}=\dfrac{5}{2} \\

& \Rightarrow 2{{x}^{2}}-5x+2=0 \\

& \Rightarrow \left( x-2 \right)\left( 2x-1 \right)=0 \\

\end{align}$

So, we get the values of x here as,

$\begin{align}

& x=2,x=\dfrac{1}{2} \\

& \Rightarrow x=\left\{ \dfrac{1}{2},1,2 \right\} \\

\end{align}$

**So, the correct option is A.**

**Note:**While solving these types of questions you should always know about the solutions of quadratic equations. These are solved by an easy method which is factorisation method and that’s why the values can be found easily. If we do not get the proper factors of the quadratic equation then we could also use the quadratic formula which is one of the methods to solve the quadratic equation equation easily.

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