The number of real solutions of the equation 3(x+1x)^2-2(x+1x)-1=0, is

Question

The number of real solutions of the equation $3\left(x+\dfrac1x\right)^2-2\left(x+\dfrac1x\right)-1=0$, is

Accepted Answer

Let $t=x+\dfrac1x$. The equation becomes $3t^2-2t-1=0 \implies (t-1)(3t+1)=0 \implies t=1 	ext{ or } t=-\dfrac13$.

For real $x
eq0$, $\left|x+\dfrac1xight|\ge2$ always. Since neither $t=1$ nor $t=-\dfrac13$ satisfies this, there are no real solutions.

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