Math help?

Question

Solve by using the most apropiate method. Give irrational roots in simplest form. (x+3)^2 + 6(x+3)=16?

Answer 1

you will square the (x+3) first and get

x^2 + 6x +9 + 6x+18 = 16 add up like terms

x^2 +12x + 27 =16 subtract the 16 on both sides

x^2 +12x + 11 = 0

factor by looking for 2 numbers that multiply to give 11 and add up to give 12 and it will be ( 1 and 11)

so you will have

(x+1)(x+11)=0 solve for x and you will end up with

x= -1 or x = -11

Answer 2

There is no need to square the binomials in this situation. Treat each (x + 3) as a single variable and factor normally:

(x+3)^2 + 6(x+3) = 16

(x+3)^2 + 6(x+3) - 16 = 0

[(x + 3) + 8] [(x + 3) - 2] = 0

(x + 3) + 8 = 0
x + 11 = 0
x = -11

(x + 3) - 2 = 0
x + 1 = 0
x = -1

There are no irrational roots in this situation, both roots (-11 and -1) are rational.

Answer 3

Let y = x + 3
y² + 6y - 16 = 0
(y + 8).(y - 2) = 0
y = - 8, y = 2
x + 3 = - 8, x + 3 = 2
x = - 11, x = - 1

Answer 4

x^2 + 6x + 9 + 6x + 18 - 16 = 0
x^2 + 12x +11 = 0
(x + 11)(x + 1) = 0
x = -11 or x = -1

Answer 5

6x + 18 + x^2 + 6 = 16

x^2 + 6x + 22 = 16

x^2 + 6x = -6

x^2 + 6x + 9 = 3

(x + 3)^2 = 3

x+3 = + and - radical 3

x = -3 + and - radical 3

Answer 6

(4+3)^2+6(6+3)=ya mamma

Answer 7

-1