solve this equation for $x$ : $y=x-6\sqrt{x}$ solve for $x$ this equation :
$$y=x-6\sqrt{x}$$
I've tried raising everything to the power of two but it doesn't work 
$x$ shouldn't have two values.
 A: $$\begin{align} y = x-6\sqrt x & \iff (y-x) = -6\sqrt x \\ \\&\implies  y^2 - 2xy + x^2 = 36x\\ \\ &\iff x^2 - (2y - 36)x + y^2 = 0\end{align}$$
Now you can solve for $x$ using the quadratic equation (treat $y$ as a constant), but you'll want to check whether both solutions work, because by squaring, we may have introduced a solution that is not valid in the original equation.
$$x = \dfrac{(2y-36) \pm \sqrt{(2y - 36)^2 - 4y^2}}{2}$$
Now, it's just simplifying, and remember to check both solutions: Recall that $x\geq 0$ or else $\sqrt x$ is not defined.
A: $$6\sqrt x=x-y$$
$$36x=x^2-2xy+y^2$$
$$x^2-(2y+36)x+y^2=0$$
$$x=\frac{2y+36\pm\sqrt{4y^2+144y+1296-4y^2}}2=y+18\pm6\sqrt{9+y}$$
Since $6\sqrt x\ge 0$ it must be $x\ge y$. Furthermore, $y\ge -9$. 
For $y=-9$ we have $x=9$. For $-9< y\le0$ there are two solutions. For $y>0$ the solution obtained choosing '$-$' in $\pm$ implies that $x-y<0$ and it is not valid.
A: $$
y= x-6\sqrt{x}
$$
$$
y = u^2 - 6u
$$
$$
u^2-6u - y = 0
$$
This last is a quadratic equation in the variable $u$.  You can use the usual formula or you can complete the square, thus:
$$
u^2-6u + 9 = y+9
$$
$$
(u-3)^2 = y+9
$$
$$
u-3 = \pm\sqrt{y+9}
$$
$$
u = 3\pm\sqrt{y+9}
$$
$$
\sqrt{x} = 3\pm\sqrt{y+9}
$$
$$
x = \cdots
$$
A: You can substitute with $u=\sqrt{x}$ to get
$$ y=u^2-6u $$
$u$ can be expressed as
$$ u = \sqrt{x} = 3 \pm \sqrt{9+y} $$
from which you can express $x$ as
$$ x = \left( 3 \pm \sqrt{9+y} \right)^2 $$
A: The substitution $u=\sqrt{x}$ is surely a good tool:
$$
y=u^2-6u
$$
becomes
$$
u^2-6u-y=0
$$
which is quadratic in $u$. Remember, though, that $u\ge0$ by definition.
If $y>0$, the quadratic has one positive and one negative solution, so we can consider only the positive one:
$$
u=3+\sqrt{9+y}
$$
and $x=(3+\sqrt{9+y})^2$.
If $y<0$, the equation has two positive solutions provided the discriminant is positive; the discriminant is $9+y$, so we get the condition $y\ge-9$ and the two solutions are
$$
x=(3-\sqrt{9+y})^2,\qquad x=(3+\sqrt{9+y})^2
$$
For $y=-9$ there's actually just one solution.
In summary:


*

*No solution for $y<-9$

*One solution for $y=-9$, namely $x=9$;

*Two solutions for $-9<y<0$, namely $x=(3-\sqrt{9+y})^2$ and $x=(3+\sqrt{9+y})^2$

*Two solutions for $y=0$, namely $x=0$ and $x=36$;

*One solution for $y>0$, namely $x=(3+\sqrt{9+y})^2$

