1
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ Hi Hubert. Welcome to MSE. Can you please format your question using LaTeX? $\endgroup$ – Pantelis Sopasakis May 8 at 19:56
3
$\begingroup$

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 $x=16$.

$\endgroup$
  • $\begingroup$ Thanks, that's what I couldn't figure out. $\endgroup$ – Hubert Hanc May 8 at 20:15
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.