How do you solve $(sqrtx) -1= sqrt(5-x)$ ?

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Answer 1 · 2017-03-24T02:16:07

$x = 4$ is applicable for +ve value of both sides
$x = 1$ is applicable for -ve value of both sides

$(sqrt x) - 1 = sqrt (5-x)$

square both sides
$((sqrt x) - 1 )^2= (sqrt (5-x))^2$

$x - 2 sqrt x + 1 = 5-x$

$x + x - 2 sqrt x + 1 -5= 0$

$2 x - 2 sqrt x - 4= 0$

Let say $sqrt x = y$ , then $x =(sqrt x)^2 = y^2$

$2 y^2 -2 y - 4 = 0$
$2(y^2 -y -2) = 0$
$2(y - 2)(y + 1) = 0$
$y = 2, y = -1$

When $y = 2$ , $x = 2^2 = 4$
When $y = -1$ , $x = (-1)^2 = 1$

How do you solve (sqrtx) -1= sqrt(5-x) (sqrtx) -1= sqrt(5-x)?