Question

How do you solve `(5n - 1) (n+1) = 0`?

Answer 1

`n=1/5` `or n=-1`

Explanation:

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

`5n^2+4n-1=0`

`(5n-1)=0`

`5n=1`

`n=1/5`

or

`(n+1)=0`

`n=-1`

substitute `n=1/5`

`5(1/5)^2+4(1/5)-1=0`

`5(1/25)+4/5-1=0`

`5/25+4/5-1=0`

`1/5+4/5-1=0`

`5/5-1=0`

`1-1=0`

substitute `n=-1`

`5n^2+4n-1=0`

`5(-1)^2+(4(-1)-1=0`

`5+(-4)-1=0`

`1-1=0`

Answer 2

`n = 1/5" or "n=-1`

Explanation:

This quadratic is already in factor form.

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

If the product of 2 factors is 0, one of them must be 0.

Consider each factor equal to 0.

If `5n-1=0color(white)(..................) "if " n+1=0`

`rarr" "5n = 1color(white)(..............)rarr" "n=-1`

`rarr" "n = 1/5`