# 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.

• Hint: Take the term with the square root to one side and square both sides. You'll get a quadratic equation. – user_of_math Nov 23 '14 at 16:07
• You say that $x$ shouldn't have $2$ values. First, this is an equation in $2$ variables so $x$ can take an infinite number of values. But maybe you mean that $x$ should only take $1$ value for every value of $y$? That would mean that this function is one-to-one. Take a look at the plot -- it's not. In fact I can easily see that $x=25$ and $x=1$ both correspond to the same $y$ value. – user137731 Nov 23 '14 at 16:15

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 $x=(3+\sqrt{9+y})^2$
• why is there only one solution for 0<y ? – Lynn Nov 23 '14 at 18:49
• @Lynn Since $u=\sqrt{x}$, you have $u\ge0$. However, an equation of the form $u^2+bu+c=0$ where $c<0$ has one positive and one negative roots (because the product of the roots is $c<0$): the negative root must be discarded. – egreg Nov 23 '14 at 19:04

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

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

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

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

• If $y=7$, then you should get two solutions, $x=1$ and $x=49$. But $1-6\sqrt{1}=-5\ne7$. – egreg Nov 23 '14 at 16:29
• $\sqrt{1}=\pm 1$ therefore $1-6(-1)=7$ – bkosztin Nov 23 '14 at 16:54
• Really? No, sorry, this is not in my math books. I always have $\sqrt{1}=1$. – egreg Nov 23 '14 at 16:55
• Every positive number a has two square roots. – bkosztin Nov 23 '14 at 17:01
• And you choose it at random which one? Please! – egreg Nov 23 '14 at 17:02

\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.

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

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.

• You should check for extraneous roots since you squared both sides. – Michael Hardy Nov 23 '14 at 16:14
• i think you forgot the x in the third step .I believe 2y+36 by -x not -.Am i wrong?please answer – Lynn Nov 23 '14 at 16:36