How do you solve $sqrt x = x - 6$ ?

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Answer 1 · 2016-07-22T23:25:27

From the relation $sqrtx=x-6$ we have that $x>=0$ and

$x-6>=0=>x>=6$ .Finally solutions must be in $[6,+oo)$

hence

$sqrtx=x-6$

$(sqrtx)^2=(x-6)^2$

$x=(x-6)^2$

$x=x^2-12x+36$

$x^2-12x-x+36=0$

$x^2-13x+36=0$

$(x-4)*(x-9)=0$

$x_1=4$ and $x_2=9$

The only acceptable solution is $x=9$

How do you solve sqrt x = x - 6sqrt x = x - 6?