Question

solve for \(x\)

Answer 1

\(\large \color{black}{\begin{align} (x-1)\sqrt{x^2-x-2}\geq 0\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

i got

\(\large \color{black}{\begin{align} x\geq 1 \cup \left(x\geq 2 \cup x\leq -1\right)\hspace{.33em}\\~\\
\end{align}}\)

Answer 3

\[x^2-x-2=(x-2)(x+1) \\ (x-1) \sqrt{(x-1)(x-2)} \ge 0\]
we should consider the domain of the square root function there
on that square root function's domain the square root function will be positive
so then you just need to consider when (x-1) is positive
and find the intersection of these sets

Answer 4

how to solve this

\(\large \color{black}{\begin{align} x\geq 1 \cup \left(x\geq 2 \cup x\leq -1\right)\hspace{.33em}\\~\\
\end{align}}\)

Answer 5

is that one symbol is suppose to be an intersect sign ?

Answer 6

i havent studied intersection of sets

Answer 7

i just tried and posted

Answer 8

we should also go back to include when the function is 0


but I think you mean
\[(x \ge 1) \cap (x \ge 2 \cup x \le -1)\]
draw a number line
in keep in mind that x can be 1,-1,2 since that makes the function 0

but anyways it might helped to also graph that...