Answer
$x\displaystyle \in\left\{ \frac{1}{2},\frac{-1}{2}, \sqrt 2i, -\sqrt 2i\right\}$
Work Step by Step
See Descarte's rule of sign:
It asserts that the number of real roots is at most the number of sign changes in the sequence of polynomial's coefficients (omitting the zero coefficients), and that the difference between these two numbers is always even.
See The Rational Zero Theorem:
If $\frac{p}{q}$ is a zero of the polynomial $f(x) $ with integer coefficients, then $p$ is a factor of the constant term, $a_{0}$, and
$q$ is a factor of the leading coefficient, $a_{n}$.
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$f(x)=4x^4+7x^{2}-2$
a. Candidates for zeros, $ \frac{p}{q}:$
$p:\qquad \pm 1, \pm2$
$q:\qquad \pm 1, \pm 2, \pm4$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm2, \pm\frac{1}{2},\frac{1}{4}$
b. $f(x)=4x^4+7x^{2}-2$, there's one sign change therefore $f(x)$ has $1$ positive root.
$f(-x)=4x^4+7x^{2}-2$, there's one sign change therefore $f(x)$ has $1$ negative root.
c. Try for $x=\frac{1}{2}:$
$\begin{array}{lllll}
\underline{\frac{1}{2}}| & 4 & 0& 7 & 0 & -2\\
& & 2 & 1 & 4&2\\
& -- & -- & -- & --\\
& 4 & 2 & 8&4 & |\underline{0}
\end{array}$
$\frac{1}{2}$ is a zero,
$f(x)=(x-\frac{1}{2})(4x^3+2x^{2} +8x+4)$
Try for $x=\frac{-1}{2}:$
$\begin{array}{lllll}
\underline{\frac{-1}{2}}| & 4 & 2 & 8 & 4\\
& & -2 & 0&-4\\
& -- & -- & -- & --\\
& 4 & 0&8& |\underline{0}
\end{array}$
$\frac{-1}{2}$ is a zero,
$f(x)=(x-\frac{1}{2})(x+\frac{1}{2})(4x^2+8)$
d. $(x-\frac{1}{2})(x+\frac{1}{2})(4x^2+8)=0$
The roots of $4x^2+8=0$ are:
$x=-\sqrt 2i,x=\sqrt 2i$
$x\displaystyle \in\left\{ \frac{1}{2},\frac{-1}{2}, \sqrt 2i, -\sqrt 2i\right\}$
