help with simple algebra equation?

Question

Hi, this problem is pretty simple but I want to make sure im doing it right.:

x(to the second power)-117=4

help please?

thank you!

Answer 1

x² - 117 = 4
x² = 117 + 4
x² = 121
x' = 11
x'' = -11

Answer 2

x = 11

Answer 3

x=11

Answer 4

x=11

Answer 5

x^2 - 117 = 4
bring the 117 to the right and it becomes positive ( or adding 117 to both sides )
it will become:
x^2 = 4 + 117
......= 121
square root both sides:
x = {square root} 121
...= positive 11 and negative 11

because when (-11)^2 = positive and (+11)^2 = also positive

Answer 6

x^2-117=4
x^2-117(+117)=4(+117)
x^2=121

so to "undo" the square take the square root of both sides of the equation (x^2, 121) which gives you:

x=+/-11

*IMPORTANT*

if your instructions define x as a positive real number then your answer will be 11, if x is any real number then you must write your answer as "+/-11".

Answer 7

x^2 - 117 = 4
x^2 = 121
square root of 121
x = 11

Answer 8

x^2-117=4
first add 117 to both sides of the equation.
x^2=121
then take the square root of both sides.
x=11

Answer 9

x² - 117 = 4
x² = 121
x = ±√121
x = ±11

So x = 11 or -11

Alternate method:

x² - 117 = 4
x² - 121 = 0
(x + 11)(x - 11) = 0
x = -11 or x = 11

Answer 10

Add 117 to both sides

x^2=121

Take square root
x=11 or -11

Answer 11

You need to add 117 to each side
xsquared= 121
then find the square root of 121. I don't have a calculator near me and am to lazy to go get one. But I think you can figure it out

x= the square root of 121