How to find value of $x?$

Question

$A = x^{1/4} $

$B = x^{1/6}$

And $A^2 = 4B$

Find values of $x$ .

Can I get a hint on how to solve this ?

Answer 1

According to given equation:

$$A^2=4B$$

Put the value of $A$ and $B$ in above equation, then: $$(x^{1/4})^2=4(x^{1/6})$$

$$(x^{1/2})=4(x^{1/6})$$

Take square both sides, then: $$x=16(x^{1/3})$$

$$(x^{2/3})\times (x^{1/3})=16(x^{1/3})$$

$$(x^{2/3})=16$$

Take square root both sides, then: $$(x^{1/3})=4$$

Take cube both sides, then: $$x=4\times4\times4=64$$

Answer 2

We are given the values \begin{align*} A & = x^{1/4}\\ B & = x^{1/6} \end{align*} and asked to solve the equation $$A^2 = 4B$$
Substituting $(x^{1/4})^2 = x^{1/2}$ for $A^2$ and $x^{1/6}$ for $B$ in the equation $A^2 = 4B$ yields $$x^{1/2} = 4x^{1/6}$$ If we make the substitution $u = x^{1/6}$, we obtain $$u^3 = 4u$$ which can be solved by subtracting $4u$ from each side of the equation, then factoring. Finally, note that $u = x^{1/6}$ denotes the principal (non-negative) real root of $x$, so $u \geq 0$.