Simplifying $a = \dfrac{\sqrt{x}}{x+3} $

Question

Solving equations involving terms of the form $ \dfrac{3x}{6x^2} $ is easy. You can cancel the $x$ in the numerator and end up with: $ \dfrac{3}{6x} $. However, I am presented with an equation of the following form: $$a = \dfrac{\sqrt{x}}{x+3} $$ Where, $a$ is a constant.

Trying to cancel out the variable $x$ in either the numerator or denominator doesn't work. How do I go about and solve such an equation?

Answer 1

$ax+3a=\sqrt{x} \implies a^2x^2+(6a^2-1)x+9a^2=0$ $$\begin{align*}\implies x&=\frac{1-6a^2\pm\sqrt{(6a^2-1)^2-36a^4}}{2a^2}\quad(a\neq 0)\\ &=\frac{1-6a^2\pm\sqrt{1-12a^2}}{2a^2}\quad(a\neq 0)\end{align*}$$

Answer 2

$$a=\frac{\sqrt x}{x+3}\quad (\forall \ \ \ x\ge 0)$$ $$ax-\sqrt x+3a=0$$ $$a(\sqrt x)^2-\sqrt x+3a=0$$ Solving above quadratic equation for $\sqrt x$ as follows $$\sqrt x=\frac{-(-1)\pm\sqrt{(-1)^2-4\cdot a\cdot 3a}}{2\cdot a}$$ $$x=\frac{(1\pm\sqrt{1-12a^2})^2}{4a^2}$$ $$\forall \ \ 0\le a\le \frac{1}{2\sqrt3}$$

Answer 3

One should be careful with simplifications and with squaring. For instance, the equation $x=1$ has just one solution, but $x^2=1^2$ has two.

Since $x=(\sqrt{x}\,)^2$, you can set $y=\sqrt{x}$ and try to solve $$ \frac{y}{y^2+3}=a $$ with the limitation that $y\ge0$ (which holds by definition of square root).

Now, before going on, we observe that there is just the solution $y=0$ (so $x=0$) when $a=0$. There is no solution for $a<0$, because the left-hand side is $\ge0$ for every $y\ge0$.

So we proceed to solve the equation with the additional assumption that $a>0$; in particular $y=0$ is not a solution.

Removing the denominator and reordering we obtain $ay^2-y+3a=0$. We note that the sum of the roots is $3/a>0$ and the product is $3a/a=3>0$. Thus the equation has two positive roots (possibly coincident) as soon as its discriminant is $\ge0$: thus the condition is $$ \Delta=1-12a^2\ge0 $$ Under this condition you can write the solutions as $$ y=\dfrac{1-\sqrt{\Delta}}{2a}\qquad\text{or}\qquad y=\dfrac{1+\sqrt{\Delta}}{2a} $$ and you can square these to find the solutions for $x$.

In conclusion: the equation has no solution for $a<0$; one solution for $a=0$; two solutions for $0<a<1/\sqrt{12}$; one solution (with multiplicity two) for $a=1/\sqrt{12}$; no solution for $a>1/\sqrt{12}$.