find the value of x

Posted January 19th, 2008 by nikimiki

$x+1=\sqrt[2]{1-x}=\sqrt{1-x}$

Comments

On January 19th, 2008 Structure says:

You want
A

$x+1=\sqrt{1-x}$

or B

$x+1=\sqrt{2(1-x})$

On January 19th, 2008 nikimiki says:

My question is x+1=\sqrt{2}(1-x)

On January 19th, 2008 Structure says:

I have changed your post to help you using latex editor.
Try to look to the format and post your question in latex

$x+1=\sqrt[2]{1-x}=\sqrt{1-x}$

$x^2+2x+1=1-x$

$x^2+3x=0$

$x=0$

The other solution

$x=-3$

of tle later equation is not a solution of the initial equation as you have to keep

$x+1\geq 0$