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

`(-18)/(128)= (128n^2)/(128)`

`(-9)/(64)= {n^2}`

`n=\sqrt{(-9)/(64)}`

`n = (3i)/(8)` and
`n= (-3i)/(8)`

So, the values of 'n' are
`n = (3i)/(8)` and
`n= (-3i)/(8)`.