# Imposter solution for$\sqrt{2x+15}-6=x$

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.

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?