Question

How do you solve `25v^2=1`?

Answer 1

`v=+-1/5`

Explanation:

Divide both sides by 25 giving:

`v^2=1/25`

Thus:

`sqrt(v^2)=sqrt(1/25)`

`v=sqrt(1)/(sqrt(25)) = 1/sqrt(25)`

But as this is square root we should write `=+-1/sqrt(25)`

`v=+-1/5`

Answer 2

`v=+-1/5`

Explanation:

Begin by isolating `v^2`. To do this we divide both sides of the equation by 25.

`rArr(cancel(25)^1 v^2)/cancel(25)^1=1/25rArrv^2=1/25`

now take the `color(blue)"square root of both sides"`

`rArrsqrt(v^2)=+-sqrt(1/25)`

`rArrv=+-1/5`

Answer 3

`v = +-1/5`

Explanation:

Apart from the method shown by other contributors, we can also follow the usual method for a quadratic equation.

`rarr "Make it equal to 0 " rarr " find the factors."`

`25v^2 = 1`

`25v^2 -1 = 0 larr" difference of squares"`

`(5v+1)(5v-1) = 0`

Either factor could be 0

`5v+1 = 0 " " rarr v = -1/5`

`5v-1 = 0 " " rarr v = 1/5`