1. ## finding de value of x

I tried to solve thees but is useless I not doing it right. Can some one lead the way?

1) sqrt(2x + 1) = x-1

2) -2< x+2/x-1

2. Originally Posted by cristinaivelisse
I tried to solve thees but is useless I not doing it right. Can some one lead the way?

1) sqrt(2x + 1) = x-1

2) -2< x+2/x-1

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

square both sides ...

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

$0 = x^2 - 4x$

$0 = x(x-4)$

$x = 0 , x = 4$

check for extraneous solution(s)

$-2 < \frac{x+2}{x-1}$

$0 < 2 + \frac{x+2}{x-1}$

$0 < \frac{2(x-1)}{x-1} + \frac{x+2}{x-1}$

$0 < \frac{3x}{x-1}$

critical values are $x = 0$ and $x = 1$. these two values of x break up the set of all x-values into three sections . ..

$x < 0$ , $0 < x < 1$ , and $x > 1$

pick any number in each section ... if it makes the original inequality true, then all values of x in that section will be in the solution set.