Where did the 1/2 go in this question? 1/2 -->(x^2-2x-3)<--square root?

Question

It goes like this:
1/2 //// (x^2-2x-3)=0
x^2-2x-3=0
(x-3)(x+1)=0
x = 3, x = -1
(3,0) (-1,0)

Where did 1/2 in front of the square root go?

Answer 1

1/2 √(x² - 2x - 3) = 0

Multiply both sides by 2:

√(x² - 2x - 3) = 2*0 = 0

I think you can see the rest from there.

Answer 2

1/2 CAN'T BE EQUAL ZERO.SO YOU FORGET ABOUT IT.

Every body pay attention :

When you have some values being multiplied and the results = 0

Every one of them could be Zero:
Example: (x-1) (x+2) = 0 , x-1 =0 or x+2 =0

Now if we had: 2(x-1)(x+2) = 0 damn, 2 can't be 0 can it ?
which means the other 2 could be equal zero.
so you forget about it? what is this multiplying or in this case dividing both side by 2 s - - t? come on guys? Learning all the times yes?

Best Regards.

Answer 3

Multiply both sides with 2.

Answer 4

no where. the first step was to multiply both sides by 2.