If $x^4 + \dfrac{1}{x^4} = \dfrac{-16}{27}$, find the value of $x+\dfrac{1}{x}$

It is given in this problem that $x^4 + \dfrac{1}{x^4}$ is equal to $\dfrac{-16}{27}$

$x^4 + \dfrac{1}{x^4} = \dfrac{-16}{27}$

The value of $x+\dfrac{1}{x}$ can be found by reducing the power of the given equation. In order to do it, express $x$ raised to the power of $4$ in terms of $x$ squared.

${(x^2)}^2 + \dfrac{1}{{(x^2)}^2} = \dfrac{-16}{27}$

$\implies {(x^2)}^2 + {\Bigg(\dfrac{1}{x^2}\Bigg)}^2 = \dfrac{-16}{27}$

The two terms of the left hand side equation are in square form. So, they can used to convert them as whole square of sum of terms. In order to convert it, add $2$ to left hand side equation and subtract it by same number.

$\implies {(x^2)}^2 + {\Bigg(\dfrac{1}{x^2}\Bigg)}^2 + 2 -2 = \dfrac{-16}{27}$

We know that the product of numbers $2$ and $1$ is equal to $2$. So, the number $2$ can be express as the product of the numbers $2$ and $1$.

$\implies {(x^2)}^2 + {\Bigg(\dfrac{1}{x^2}\Bigg)}^2 + 2 \times 1 -2 = \dfrac{-16}{27}$

In L.H.S, $x^2$ and $\dfrac{1}{x^2}$ are two terms in square form. The product of them is equal to $1$. Therefore, write the number $1$ as the division of $x^2$ by $x^2$ term.

$\implies {(x^2)}^2 + {\Bigg(\dfrac{1}{x^2}\Bigg)}^2 + 2 \times \dfrac{x^2}{x^2} -2 = \dfrac{-16}{27}$

Take $x^2 = a$ and $\dfrac{1}{x^2} = b$.

$\implies a^2 + b^2 + 2ab -2 = \dfrac{-16}{27}$

The first three terms of the left hand side equation is the expansion of the a plus b whole square formula.

$\implies {(a + b)}^2 -2 = \dfrac{-16}{27}$

$\implies {(x^2)}^2 + {\Bigg(\dfrac{1}{x^2}\Bigg)}^2 + 2 \times x^2 \times \dfrac{1}{x^2} -2 = \dfrac{-16}{27}$

Replace the assumed values of $a$ and $b$ terms.

$\implies {\Bigg(x^2 + \dfrac{1}{x^2}\Bigg)}^2 = 2 -\dfrac{16}{27}$

$\implies {\Bigg(x^2 + \dfrac{1}{x^2}\Bigg)}^2 = \dfrac{2}{1} -\dfrac{16}{27}$

$\implies {\Bigg(x^2 + \dfrac{1}{x^2}\Bigg)}^2 = \dfrac{54 -16}{27}$

$\implies {\Bigg(x^2 + \dfrac{1}{x^2}\Bigg)}^2 = \dfrac{38}{27}$

Take square root both sides.

$\implies \sqrt{{\Bigg(x^2 + \dfrac{1}{x^2}\Bigg)}^2} = \sqrt{\dfrac{38}{27}}$

Once again, use the same above procedure to reduce the power of square form terms.

$\implies x^2 + \dfrac{1}{x^2} = \sqrt{\dfrac{38}{27}}$

$\implies x^2 + \dfrac{1}{x^2} + 2 -2 = \sqrt{\dfrac{38}{27}}$

$\implies x^2 + \dfrac{1}{x^2} + 2 \times 1 -2 = \sqrt{\dfrac{38}{27}}$

$\implies x^2 + \dfrac{1}{x^2} + 2 \times \dfrac{x^2}{x^2} -2 = \sqrt{\dfrac{38}{27}}$

$\implies x^2 + \dfrac{1}{x^2} + 2 \times x^2 \times \dfrac{1}{x^2} -2 = \sqrt{\dfrac{38}{27}}$

$\implies x^2 + \dfrac{1}{x^2} + 2 \times x^2 \times \dfrac{1}{x^2} = 2 + \sqrt{\dfrac{38}{27}}$

$\implies {\Bigg(x + \dfrac{1}{x}\Bigg)}^2 = 2 + \sqrt{\dfrac{38}{27}}$

$\implies \sqrt{{\Bigg(x + \dfrac{1}{x}\Bigg)}^2} = \sqrt{2 + \sqrt{\dfrac{38}{27}}}$

$\therefore \,\,\,\,\, x + \dfrac{1}{x} = \sqrt{2 + \sqrt{\dfrac{38}{27}}}$

Therefore, the value of $x + \dfrac{1}{x}$ is equal to $\sqrt{2 + \sqrt{\dfrac{38}{27}}}$