Question

Find all the real fourth roots of 256/2401

Answer 1

Answer:

Real roots of $\frac{256}{2401}$ are ${\frac{4}{7}}$ and ${\frac{-4}{7}}$

Step-by-step explanation:

We have to find the fourth root of $\frac{256}{2401}$

that is $x^4=\frac{256}{2401}$

$\Rightarrow x^4-\frac{256}{2401}=0$

Using identity , $a^2-b^2= (a+b)(a-b)$

$\Rightarrow (x^2-\frac{16}{49})(x^2+\frac{16}{49})=0$

Using zero product rule, it states if product of two number is zero then either first number is zero or second number is zero that is, $a.b=0 \Rightarrow a=0 \ \text{or} \ b=0$

thus,

$\Rightarrow (x^2-\frac{16}{49})=0$ or $\Rightarrow (x^2+\frac{16}{49})=0$

$\Rightarrow x^2=\frac{16}{49}$ or $\Rightarrow x^2=\frac{-16}{49}$

$\Rightarrow x^2=\frac{-16}{49}$ this will complex roots.

$\Rightarrow x^2=\frac{16}{49}$

$\Rightarrow x=\sqrt{\frac{16}{49}}$

$\Rightarrow x=\pm{\frac{4}{7}}$

Thus, real roots of $\frac{256}{2401}$ are ${\frac{4}{7}}$ and ${\frac{-4}{7}}$

Answer 2

The real fourth roots of  256/2401 are 4/7 and -4/7. The other two roots are complex. Hope this helps and have a wonderful day...;)