Question

Find all the real fourth roots of $\frac{256}{2401}$.

Answer 1

Finding the real roots of ${\left(\frac{256}{2401}\right)}^{\frac{1}{4}}$

Let us write,

$$\begin{gathered} x={\left(\frac{256}{2401}\right)}^{\frac{1}{4}} \\ \Rightarrow x^4=\frac{256}{2401} \end{gathered}$$

On factorization of the given fraction, we can write

$\frac{256}{2401}={\left(\frac{16}{49}\right)}^2={\left(\frac{4}{7}\right)}^4$

Therefore,

$\begin{array}{rcl}{\mathrm{x}}^4& =& {\left(\frac{4}{7}\right)}^4\\ & \Rightarrow & {\mathrm{x}}^4-{\left(\frac{4}{7}\right)}^4=0\\ & \Rightarrow & \left({\mathrm{x}}^2-{\left(\frac{4}{7}\right)}^2\right)\left({\mathrm{x}}^2+{\left(\frac{4}{7}\right)}^2\right)=0\left[\because \left({\mathrm{a}}^2-{\mathrm{b}}^2\right)=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)\right]\\ & \Rightarrow & \left(\mathrm{x}-\frac{4}{7}\right)\left(\mathrm{x}+\frac{4}{7}\right)\left({\mathrm{x}}^2+\frac{16}{49}\right)=0\\ & \Rightarrow & \mathrm{x}=\frac{4}{7},-\frac{4}{7}\end{array}$

Hence, the real fourth roots of $\frac{256}{2401}$ is $\frac{4}{7}\&-\frac{4}{7}$.