How do you solve #3( x + 1)^{2} - 5= 7#?

1 Answer
Oct 11, 2016

The solutions of the equation are #x = -3# and #x = 1#.

Explanation:

Begin by expanding the binomial squared and completing that multiplication. Then distribute the #3#. Combine like terms on the left side of the equation. Get all terms on the left side of the equation, equal to #0# on the right side of the equation. Since you have a quadratic equation, to find its two solutions, you must factor the equation and apply the Zero Product Property to find the solutions from the factors.

#3(x + 1)^2 - 5 = 7#
#3(x + 1)(x + 1) - 5 = 7#
#3(x^2 + x + x + 1) - 5 = 7#
#3(x^2 + 2x + 1) - 5 = 7#
#3x^2 + 6x + 3 - 5 = 7#
#3x^2 + 6x - 2 = 7#
#3x^2 + 6x - 2 - 7 = 7 - 7#
#3x^2 + 6x - 9 = 0#

Now reduce and factor the equation.
#(3x^2 + 6x - 9)/3 = 0/3#
#x^2 + 2x - 3 = 0#
#(x + 3)(x - 1) = 0#

Now solve each factor.
#x + 3 = 0#
#x + 3 - 3 = 0 - 3#
#x = -3#
And
#x - 1 = 0#
#x - 1 + 1 = 0 + 1#
#x = 1#