Question

What is the value of n in the equation 0 = 18 + 128n2

Answer 1

Consider the equation:

$0= 18+128n^2$

Subtracting '18' from both the sides of the equation, we get

$0-18 = 18+128n^2-18$

$-18 = 128n^2$

Dividing by '128' from both the sides of the equation, we get

$\frac{-18}{128}= \frac{128n^2}{128}$

$\frac{-9}{64}= {n^2}$

$n=\sqrt{\frac{-9}{64}}$

$n = \frac{3i}{8}$ and $n= \frac{-3i}{8}$

So, the values of 'n' are $n = \frac{3i}{8}$ and $n= \frac{-3i}{8}$.