3
$\begingroup$

I was reading this post and came across a similar but different question. Why does $\sqrt{2x+15}-6=x$ have an "imposter" solution?.

With the normal solution which involves squaring both sides, I can see how an extra solution is generated. But if I solve it with the following way:

$\sqrt{2x+15} - 6= x$ ($1$)

let $\sqrt{2x+15} + 6= y$

Then $xy=2x+15-36=2x-21$

Hence $y={2x-21\over x}$

Now $\sqrt{2x+15} + 6 ={2x-21\over x}$ ($2$)

Substract (2) with (1) and we get $12={2x-21\over x}-x$

Multiplying $x$ on both side we get $x^2+10x+21=0$ so a imposter solution is still generated.

I am wondering where exactly the $-7$ comes from in this solution. Both ($1$) and ($2$) does not accept $-7$ as a valid solution but the substraction seems to cause this extra solution to be introduced. But we are not squaring anything, a simple substraction causing a redundant root seems completely unreasonable to me.

Any idea will be appreciated.

$\endgroup$
2
$\begingroup$

Consider this: $$\begin{align} &(1)\ \ \ x-6 &=14 \\ &(2)\ \ \ x+6 &=26 \end{align}$$ Observe that the only solution to this system of equation is $x=20$. However if you subtract (1) from (2) you'd get $$ 12=12 $$ Which admits ANY value of $x$ as a solution. Where do you think all other impostor solutions come from?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you. I now understand where it comes from now. $\endgroup$ – cr001 Oct 9 '15 at 10:05
  • $\begingroup$ You're welcome :) $\endgroup$ – BigbearZzz Oct 9 '15 at 10:06

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.