How do you solve #(x + 3) (x + 4) = 5#?

1 Answer
Aug 5, 2015

Answer:

#x_(1,2) = (-7 +-sqrt(21))/2#

Explanation:

Start by using the FOIL method to expand the two parantheses

#(x+3)(x+4) = x^2 + 3x + 4x + 12 = x^2 + 7x + 12#

Your equation now becomes

#x^2 + 7x + 12 = 5#

Rearrange so that all the terms are on one side of the equation

#x^2 + 7x + 7 = 0#

You can solve this quadratic equation by using the quadratic formula

#x_(1,2) = (-7 +- sqrt(7^2 - 4 * 1 * 7))/(2 * 1)#

#x_(1,2) = (-7 +- sqrt(21))/2 = {(x_1 = (-7 + sqrt(21))/2), (x_2 = (-7 - sqrt(21))/2) :}#

Check to see if these values are actual solutions

#((-7 + sqrt(21))/2 + 3) ((-7+sqrt(21))/2 + 4) = 5#

#((-7+sqrt(21))/2)^2 + 7 * ((-7 + sqrt(21))/2) + 12 = 5#

#(49 - 14sqrt(21) + 21)/4 + (-49 + 7sqrt(21))/2 = -7#

#70 - color(red)cancelcolor(black)(14sqrt(21)) - 98 + color(red)cancelcolor(black)(14sqrt(21)) = -28#

#-28 = -28# #-># #x_1# is a valid solution!

Now check for the second solution

#((-7-sqrt(21))/2)^2 + 7 * ((-7 - sqrt(21))/2) + 12 = 5#

#(49 + 14sqrt(21) + 21)/4 + (-49 - 7sqrt(21))/2 = -7#

#70 + color(red)cancelcolor(black)(14sqrt(21)) - 98 - color(red)cancelcolor(black)(14sqrt(21)) = -28#

#-28 = -28 -> x_2# is a valid solution as well!