Question

can some one please help with this. Sqrt(x+1)=x-1

Answer 1

\[\sqrt{x} ^{2} =x\]

Answer 2

is it like this?


if so square both sides and get

x+1 = (x-1)^2

(x-1)^2 = x^2 - 2x + 1
so
x^2 - 2x + 1 = x + 1
x^2 - 3x = 0
factor out x

x(x-3) = 0
so
x = 0 and x = 3

Answer 3

@jay657 yes.

Answer 4

yea so your answer should be
x = 0 and x = 3 as i stated above

Answer 5

\[\sqrt{x+1} = x-1\]

Squaring both sides:

\[x+1 = (x-1)^2\]
\[x+1 = x^2 + 1 - 2x\]
\[x^2 - 3x = 0\]
\[x(x-3) = 0\]

So, \(x = 0\) And \(x = 3\)..

Answer 6

Any problem @seattle12345 ??

Answer 7

@waterineyes I guess I didnt realize it needed to be factored. Is that because it is quadratic?