Question

Help.

Answer 1

@noodlesandriceyt

Answer 2

if you subtract 3x + 1 from both sides you end up with

x^2 - 36 = 0

which can be factored as a difference of squares

Answer 3

Difference of squares? thats when you take the Sqr?

Answer 4

a^2 - b^2 = (a+b)(a-b)

apply this logic to x^2 - 36

Answer 5

i thought you did uh \[\sqrt{-36}\]

Answer 6

and the same for x^2

Answer 7

x^2 - 36 = 0

(x+6)(x-6) = 0

therefore, x can be 6 or -6

Answer 8

Either you set it equal to 0 after that, or thats the final answer.

Answer 9

subtracting 3x + 1 from both sides lets us get 0 on the right side

factoring the left side gives us (x+6)(x-6)

if (x+6)(x-6) = 0 then you can solve for x by setting x + 6 = 0 or x - 6 = 0 to get both x values

Answer 10

Yes. Thank you!!!

Answer 11

uhhhhhhhhhh...It actually said that wasn't right.

Answer 12

I think we made a mistake somewhere?

Answer 13

did you enter both 6 and -6? (not sure how your program wants it formatted, probably separated by commas?)

Answer 14

Yes,

Answer 15

OK i wrote it wrong. Im used to writing it in parenthesis and it didn't take it for some reason. Sorry that was mb its right!