Question

what is kaurniya
$x + \sqrt{x = 12}$

Answer 1

let y^2 =x=> y^2+y=12=> we get upon factorization
$= > {y}^{2} + 4y - 3y - 12 = 0$
$= > y(y + 4) - 3(y + 4) = 0$
$= > (y - 3)(y + 4) = 0 = >$
So y=3 or -4. hence x=9 or 16 hence when we substitute x=16,16+√16=16+4=20 it is not a solution.
So x=9 is the solution of x+√x=12.