The number of real solutions of the equation [3( x^2 + 1 x^2 ) - 2( x + 1 x ) + 5 = 0], is

Question

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

Accepted Answer

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

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

Put $x+\frac{1}{x}=t \Rightarrow t \in(-\infty,-2] \cup[2, \infty)$

$$
\begin{aligned}
& 3 t^{2}-2 t-1=0 \\
& 3 t^{2}-3 t+t-1=0 \\
& \Rightarrow 3 t(t-1)+1(t-1)=0 \Rightarrow t=1,=-\frac{1}{3} \\
& \Rightarrow \quad t=1,-\frac{1}{3} \\
& \because t \in(-\infty,-2] \cup[2, \infty)
\end{aligned}
$$

No real value of $t \Rightarrow$ no real value of $x$.

Solutions · 1