solve this equation for $x$ : $y=x-6\sqrt{x}$

Question

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.

Answer 1

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 $x=(3+\sqrt{9+y})^2$

Answer 2

$$\begin{align} y = x-6\sqrt x & \iff (y-x) = -6\sqrt x \\ \\&\implies y^2 - 2xy + x^2 = 36x\\ \\ &\iff x^2 - (2y - 36)x + y^2 = 0\end{align}$$

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.

Answer 3

$$6\sqrt x=x-y$$ $$36x=x^2-2xy+y^2$$ $$x^2-(2y+36)x+y^2=0$$ $$x=\frac{2y+36\pm\sqrt{4y^2+144y+1296-4y^2}}2=y+18\pm6\sqrt{9+y}$$

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.

Answer 4

$$ y= x-6\sqrt{x} $$ $$ y = u^2 - 6u $$ $$ u^2-6u - y = 0 $$ This last is a quadratic equation in the variable $u$. You can use the usual formula or you can complete the square, thus:

$$ u^2-6u + 9 = y+9 $$ $$ (u-3)^2 = y+9 $$ $$ u-3 = \pm\sqrt{y+9} $$ $$ u = 3\pm\sqrt{y+9} $$ $$ \sqrt{x} = 3\pm\sqrt{y+9} $$ $$ x = \cdots $$

Answer 5

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

$$ x = \left( 3 \pm \sqrt{9+y} \right)^2 $$