Question

How do you solve `(x+1)^2-3=13`?

Answer 1

`x = 3 or x = -5`

Explanation:

Although this is a quadratic equation, we do not have to use the normal method of making it equal to 0. This is a special case - there is no x-term.

Move all the constants to the right hand side and then find the square root of each side.

`(x+1)^2 = 16`

`x+1 = +-sqrt16 = +-4`

`if x+1 =+4 " " rarr x = 3`

`if x+1 = -4" " rarr x = -5`