$$5-x^2-\frac{4}{x^2} = 0$$

Answer

$$ \begin{matrix}x_1 = 1 & x_2 = -1 & x_3 = 2 \\[1 em] x_4 = -2 & \\[1 em] \end{matrix} $$

Explanation

$$ \begin{aligned} 5-x^2-\frac{4}{x^2} &= 0&& \text{multiply ALL terms by } \color{blue}{ x^2 }. \\[1 em]x^2\cdot5-x^2x^2-x^2\cdot\frac{4}{x^2} &= x^2\cdot0&& \text{cancel out the denominators} \\[1 em]5x^2-x^4-4 &= 0&& \text{simplify left side} \\[1 em]-x^4+5x^2-4 &= 0&& \\[1 em] \end{aligned} $$

$ \color{blue}{ -x^{4}+5x^{2}-4 } $ is a polynomial of degree 4. To find zeros for polynomials of degree 3 or higher we use Rational Root Test.

The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction $ \dfrac{p}{q} $, where p is a factor of the trailing constant and q is a factor of the leading coefficient.

The factor of the leading coefficient ( -1 ) is 1 .The factors of the constant term (-4) are 1 2 4 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 4 }{ 1 } ~ $$

Substitute the POSSIBLE roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.

If we plug these values into the polynomial $ P(x) $, we obtain $ P(1) = 0 $.

To find remaining zeros we use Factor Theorem. This theorem states that if $\frac{p}{q}$ is root of the polynomial then this polynomial can be divided with $ \color{blue}{q x - p} $. In this example:

Divide $ P(x) $ with $ \color{blue}{x - 1} $

$$ \frac{ -x^{4}+5x^{2}-4 }{ \color{blue}{ x - 1 } } = -x^{3}-x^{2}+4x+4 $$

Polynomial $ -x^{3}-x^{2}+4x+4 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ -x^{3}-x^{2}+4x+4 $

When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.

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