Solution verification: solving $\sqrt{x-4}-\sqrt{x-5}+1=0$ I solved the following equation, and I just want to be sure I did it right.
This is the procedure:
$$
\sqrt{x-4}-\sqrt{x-5}+1=0\\
\sqrt{x-4}=\sqrt{x-5}-1\\
\text{squaring both sides gives me:}\\
(\sqrt{x-4})^2=(\sqrt{x-5}-1)²\\
x-4=(\sqrt{x-5})²-\sqrt{x-5}+1\\
x-4=x-5-\sqrt{x-5}+1\\
x-4=x-4-\sqrt{x-5}\\
\text{substracting x, and adding 4 to both sides}\\
0=-\sqrt(x-5)\\
\text{switching both sides}\\
\sqrt{x-5}=0\\
\text{sqaring both sides}\\
x-5=0\\
x=5\\
\text{When I place 5 in the equation, I get:}\\
\sqrt{5-4}-\sqrt{5-5}+1=0\\
\sqrt{1}-\sqrt{0}+1=0\\
1-0+1=0\\
2=0\\
\text{this means that the equation dosent have any solution, right??}\\
$$
Any advice and suggestion is helpful.
Thanks!!!
 A: Easiest way to see it, is take both square roors to the other side and square, to get
$$
\begin{split}
1 &= (x-5)-(x-4) - 2\sqrt{(x-5)(x-4)}\\
2 + 2\sqrt{(x-5)(x-4)} &=  0\\
1 + \sqrt{(x-5)(x-4)} &=  0\\
\end{split}
$$
but $\sqrt{\ldots} \geq 0$ so this is impossible...
A: Note that the expansion of $(\sqrt{x - 5} - 1)^2 = (\sqrt{x - 5})^2 - {\bf 2}\sqrt {x - 5} + 1$. 
But that correction doesn't change the fact that indeed, no solution exists.
A: It is easily seen that the equation $\sqrt{x-4}-\sqrt{x-5}=-1$ has no solutions since $\sqrt{x-4}>\sqrt{x-5}$ for all $x>=5$.
Your method is correct but you made a mistake in going from step $3$ to step $4$. After correcting that, you will still reach the same conclusion that there are no solutions.
A: The square root function is increasing on $[0,\infty)$, so
$$\sqrt{x-4} - \sqrt{x-5} \ge  0 $$
for all $x\ge 5.$  Hence the expression can never equal -1.
A: Hint $\ \ \sqrt{z+1}+\sqrt{z}\, =:\, y\ $ times $\,(\sqrt{z+1}-\sqrt{z}\,=\,-1) \ \Rightarrow\  1 = -y\,$ contra $\, y > 0\ \ \ $ QED
A: As $(x-4)-(x-5)=1$
$\displaystyle(\sqrt{x-4}-\sqrt{x-5})(\sqrt{x-4}+\sqrt{x-5})=1$
As $\displaystyle\sqrt{x-4}-\sqrt{x-5}=-1, \sqrt{x-4}+\sqrt{x-5}=-1$
Adding we get $\displaystyle\sqrt{x-4}=-1$ which is impossible as $\sqrt{x-1}\ge0$ for real $x$
A: Set $a=\sqrt{x-4}$ and $b=\sqrt{x-5}$ for $-2\leq x\leq +2$. Equivalently we have $a^2=x-4$,  $b^2={x-5}$ for $-2\leq x\leq +2$. It's implies for $-2\leq x\leq +2$,
\begin{align}
a-b=-1 \implies & (a-b)(a+b)=-1\cdot(a+b)\\
       \implies & a^2-b^2=-(a+b)\\
       \implies & (x-4)-(x-5)=-(a+b)\\
       \implies & -1=-(a+b)\\
       \implies & a+b=+1.\\
\end{align}
The same calculations and similar implications $a+b=+1\implies a-b=+1$. Then $ x $ is a root of $a+b=-1$, if only if, 
\begin{cases}
a+b=&+1\\
a-b=&-1\\
\end{cases}
Here, $a=0$ and $b=1$ implies $a^2=x-4=0$ and $b^2={x-5}=1$. In both cases, $x\in\!\!\!\!\!/[-2,+2]$.
So the equation 
$$
\sqrt{x-4}-\sqrt{x-5}+1=0 
$$
has no solution.
