The number of real solutions of the equation 3( x ^ 2 + 1 x ^ 2 ) -...

Given equation is:

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

We know,

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

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

So, we can write the given equation as

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

Let $x + \dfrac{1}{x} = t$

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

$⇒ 3{t}^{2} - 2t - 1 = 0 ⇒ 3{t}^{2} - 3t + t - 1 = 0$

$⇒ 3t\left(t - 1\right) + ( t - 1 ) = 0 ⇒ \left(3t + 1\right)\left(t - 1\right) = 0$

Therefore, $t = - \dfrac{1}{3}ort = 1$

$CaseI$

$x + \dfrac{1}{x} = 1$

$⇒ {x}^{2} - x + 1 = 0$

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

$CaseIIx + \dfrac{1}{x} = - \dfrac{1}{3}$

$⇒ 3{x}^{2} + x + 3 = 0$

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

Hence, we don't have real solutions.