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

1 Answer
Sep 6, 2016

Answer:

#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#