# How do you solve $\frac{x-1}{\sqrt{x}+2}=\frac{5}{2}$?

I solved it using the quadratic formula and it went like: $$\begin{gather} \frac{x-1}{\sqrt{x}+2}=\frac{5}{2} \\ 2(x-1)=5(\sqrt{x}+2) \\ 2x-2=5\sqrt{x}+10 \\ 2x-12=5\sqrt{x} \\ 4x^2+144-48x=25x \tag*{(squaring both sides)}\\ 4x^2-73x+144=0 \end{gather}$$ Then do the usual stuff and the solutions are $$16$$ and $$9/4$$. However $$9/4$$ doesn't work when you apply it to the initial equation. Is that solution correct, if not, where is the mistake?

• Hi Hubert. Welcome to MSE. Can you please format your question using LaTeX? – Pantelis Sopasakis May 8 at 19:56

If you rewrite the equation as $$2(x-1)=5(\sqrt{x}+2)$$ and then as $$2x-12=5\sqrt{x}$$ you can immedately notice that a necessary condition for a solution is $$x\ge6$$, because the right-hand side is positive.

With this side condition, you can square: $$4x^2-48x+144=25x$$, which indeed has the roots $$16$$ and $$9/4$$, but the latter fails to satisfy the condition.

Indeed, if you substitute it in the left-hand side you get $$9/2-12=-15/2$$, while the right-hand side is $$15/2$$.

A different strategy could be to set $$t=\sqrt{x}+2$$, with the condition $$t\ge2$$, because $$\sqrt{x}\ge0$$. Then we have $$x=t^2-4t+4$$ and the equation becomes $$t^2-4t+3=\frac{5}{2}t$$ or $$2t^2-13t+6=0$$ that has roots $$6$$ and $$1/2$$. Only the first one is good, whence $$\sqrt{x}=4$$ and $$x=16$$.

• Thanks, that's what I couldn't figure out. – Hubert Hanc May 8 at 20:15

If you do your derivation backwards (reading bottom to top) and take the square root of $$25x$$,

the previous line would be $$\pm5\sqrt x$$.

$$\dfrac94$$ is a solution of $$2x-12=-5\sqrt x$$.

That's why you always have to check that the solutions you get are valid.