Solve in R this equation :
1. Substitute $\displaystyle u = \sqrt{x-1}$
then $\displaystyle u^2 = x-1$
2. $\displaystyle \sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ becomes:
$\displaystyle \sqrt{u^2+4-4u}+\sqrt{u^2+9-6u}=1$ which is:
$\displaystyle \sqrt{(u-2)^2}+\sqrt{(u-3)^2}=1$ and you get:
$\displaystyle |u-2|+|u-3|=1~\implies~2\leq u \leq 3$
3. Now resubstitute:
$\displaystyle 2\leq \sqrt{x-1} \leq 3$
and solve for x
4. This part is for you.
That isn't a condition but the solution of the equation!
Use the definition of the absolute value: $\displaystyle |a|=\left\{\begin{array}{r}a, a \geq 0\\-a, a < 0\end{array}\right.$
1. $\displaystyle |u-2|= \left\{\begin{array}{r}u-2, u \geq 2\\-(u-2), u < 2\end{array}\right.$
2. $\displaystyle |u-3|= \left\{\begin{array}{r}u-3, u \geq 3\\-(u-3), u < 3\end{array}\right.$
3. You get three different domains of the equation:
$\displaystyle u\geq 3 : u-2+u-3=1~\implies~2u=6~\implies~u=3$
$\displaystyle 2 \leq u < 3 : u-2-u+3=1~\implies~0=0~\implies~2 \leq u < 3$
$\displaystyle u < 2 : -u+2-u+3=1~\implies~-2u=-4~\implies~u=2$
Now collect all partial solutions and you'll get: $\displaystyle 2 \leq u \leq 3$
I'm not sure what exactly you have done ...
As I wrote in my previous post:
you only have to get rid of the square-root. So square the complete inequality and you'll get:3. Now resubstitute:
$\displaystyle 2\leq \sqrt{x-1} \leq 3$
and solve for x
$\displaystyle 2\leq \sqrt{x-1} \leq 3~\implies~4\leq x-1 \leq 9~\implies~\boxed{5\leq x \leq 10}$
I've attached the graph of the function.