Question

What are the real fourth roots of $256$?

Answer 1

Finding the real fourth root of

Fourth root: When a number is multiplied by itself four times and gives the original number, it is known as the fourth root of a number.

Let $x$ be the fourth root of $256$.

$x=\sqrt[4]{256}$

$\Rightarrow$ $x^4=256$

$\Rightarrow$ $0=x^4-256$

$\Rightarrow$ $0=\left[{\left(x^2\right)}^2-{\left(16\right)}^2\right]$

$\Rightarrow$ $0=\left(x^2-16\right)\left(x^2+16\right)$ ; $\left[\because \left(a^2-b^2\right)=\left(a+b\right)\left(a-b\right)\right]$

$\Rightarrow$ $0=\left[\left(x+4\right)\left(x-4\right)\right]\left(x^2-\left(-16\right)\right)$ ; $\left[\because \left(a^2-b^2\right)=\left(a+b\right)\left(a-b\right)\right]$

$\Rightarrow$$0=\left[\left(x+4\right)\left(x-4\right)\right]\left(x^2-{\left(4i\right)}^2\right)$

$\Rightarrow$ $0=\left[\left(x+4\right)\left(x-4\right)\right]\left[\left(x+4i\right)\left(x-4i\right)\right]$ ;$\left[\because \left(a^2-b^2\right)=\left(a+b\right)\left(a-b\right)\right]$

$\Rightarrow$ $x+4=0,x-4=0,x+4i=0,x-4i=0$

$\Rightarrow$ $x=-4,x=4,x=-4i,x=4i$

$x=4i$ and $x=-4i$ are the imaginary roots and $x=4$ and $x=-4$ are the real root of $256$.

Hence, the real fourth root of $256$ are $4$ and $-4$.