Question

solve(x+2)^2=1 please help best answer will get the medal. Please helppppp!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

(x+2)^2=1
(x+2)(x+2)=1
x^2+4x+4=1
x^2+4x=3
4x=3-x^2
x=(3-x^2)\4

Answer 2

The right answer is =-3 =-1 but i need to know how to solve it.

Answer 3

(x + 2)^2 = (x + 2)(x + 2) = x^2 + 4x + 4
so you get x^2 + 4x + 4 = 1
- 1 -1
x^2 + 4x + 3 = 0
then factor so (x + 3)(x + 1) = 0
so solve for x when
x + 3 = 0 and x + 1 = 0
that would give you your answers :)

Answer 4

Hi, First you need to "get rid" of that exponent by distributing it to the parenthesis.
So, you need to:
1. Square the first term x^2
2. Multiply the two terms and double them 4x
3. square the last term 4
Making your final answer for what is in parenthesis : x^2 + 4x + 4
Set this equal to 1 (it's in the problem that way)
then subtract it, leaving the final equation as: x^2 +4x + 3 = 0
Now factor:
(x+3)(x+1)= 0
and solve for x!
Hope that helps :)

Answer 5

Glad to help!

Answer 6

thank u

Answer 7

Any time!

Answer 8

could you help with another Q

Answer 9

Sure!