Answer
please see "step by step"
Work Step by Step
By the Rational Zero Theorem, if there is a rational root $\displaystyle \frac{p}{q}$,
then p is a factor of 2 and
q is a factor of 1.
Possible roots: $\pm 1,\pm 2$
Evaluating
$f(1)= (1)^{4}+6(1)^{2}+2=1+6+2\neq 0$
$f(-1)= (-1)^{4}+6(-1)^{2}+2=1+6+2\neq 0$
$f(2)=2^{4}+6(2^{2})+2=16+24+2\neq 0$
$f(-2)=(-2)^{4}+6(-2)^{2}+2=16+24+2\neq 0$
so, none of the "candidates" is a zero of f,
meaning that f(x)=0 has no RATIONAL roots.
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ALTERNATIVELY:
$x^{4}$ and $x^{2}$ are never negative (for real numbers x).
so the expression
$x^{4}+6x^{2}+2$ is never less than $2$ (for real numbers x).
So...
$...$it is never zero (for real numbers x).
So, for no real x is $f(x)=0.$
The equation has no real roots.
It, then has no rational roots either (every rational number is real).
