The number of real solutions of the equation | vedclass

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Question

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

Vedclass · Accepted Answer

$0$

Vedclass · Answer

$3\left(x^2+\frac{1}{x^2}ight)-2\left(x+\frac{1}{x}ight)+5=0$

$3\left[\left(x+\frac{1}{x}ight)^2-2ight]-2\left(x+\frac{1}{x}ight)+5=0$

Let $x +\frac{1}{ x }= t$

$3 t ^2-2 t -1=0$

$3 t ^2-3 t + t -1=0$

$3 t ( t -1)+1( t -1)=0$

$( t -1)(3 t +1)=0$

$t =1,-\frac{1}{3}$

$x +\frac{1}{ x }=1,-\frac{1}{3} \Rightarrow 	ext { No solution. }$

The number of real solutions of the equation $3\left(x^2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+5=0$, is

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