Mathematics - Quadratic Equation Question with Solution | TestHub

Given equation is:

$3\left({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}\right)-2\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)+5=0$

We know,

${\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)}^2={\mathrm{x}}^2+ 2+\frac{1}{{\mathrm{x}}^2} \left(\text{using algebraic identities}\right)$

$\Rightarrow {\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}={\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)}^2-2$

So, we can write the given equation as

$3\left[{\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)}^2-2\right]-2\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)+5=0$

Let $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{t}$

$\Rightarrow 3\left({\mathrm{t}}^2-2\right)-2\mathrm{t}+5=0\Rightarrow 3{\mathrm{t}}^2-6-2\mathrm{t}+5=0$

$\Rightarrow 3{\mathrm{t}}^2-2\mathrm{t}-1=0\Rightarrow 3{\mathrm{t}}^2-3\mathrm{t}+\mathrm{t}-1=0$

$\Rightarrow 3\mathrm{t}\left(\mathrm{t}-1\right)+(\mathrm{t}-1)=0\Rightarrow \left(3\mathrm{t}+1\right)\left(\mathrm{t}-1\right)=0$

Therefore, $\mathrm{t}=-\frac{1}{3} \mathrm{or} \mathrm{t}=1$

$\mathrm{Case} \mathrm{I}$

$\mathrm{x}+\frac{1}{\mathrm{x}}=1$

$\Rightarrow {\mathrm{x}}^2-\mathrm{x}+1=0$

$\text{Here}, \mathrm{D}=\sqrt{-3}\text{ i.e. no real solution}$

$$\begin{gathered} \mathrm{Case} \mathrm{II} \\ \mathrm{x}+\frac{1}{\mathrm{x}}=-\frac{1}{3} \end{gathered}$$

$\Rightarrow 3{\mathrm{x}}^2+\mathrm{x}+3=0$

$\text{Here}, \mathrm{D}=\sqrt{-35} \text{i.e. no real solution}$

Hence, we don't have real solutions.