How do you solve the equation #(3x+2)^2-49=0#?

Question

4654 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-19T18:16:55

#x=5/3# and #x=-3#

#(3x+2)^2-49=0# can be written as

#(3x+2)^2-7^2=0#

As #a^2-b^2# can be factorized as #(a+b)(a-b)#, this can be factorized as

#(3x+2+7)(3x+2-7)=0#

Hence, either #3x+2+7=0# i.e. #3x=-9# and #x=-3#

or #3x+2-7=0# i.e. #3x=7-2=5# and #x=5/3#

Answer 2 · 2017-02-19T18:22:35

Alternative to get #x=5/3# and #x=-3#

You can also treat this as a completing the square problem where the step of making the square is already done for you.

Move the 49 to the other side:

#(3x+2)^2=49#

Square root both sides:

#3x+2=(+/-)7#

Treat it as two problems and solve for x

#3x+2=7#

#3x=5#

#x=5/3#

That's the first answer

#3x+2=-7#

#3x=-9#

#x=-3#

That's the second answer.