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

Question

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?

Answer 1

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 $15/2$.

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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 $x=16$.

Answer 2

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.