Question

solve the radical inequality

Answer 1

\[\sqrt{x+2} \le x-10\]\[x+2 \le (x-10)^2\]\[x+2 \le x^2 - 20x +100\]\[0 \le x^2 - 21x + 98\]
Treat as equation
\[0 = x^2 - 21x + 98\]\[x = -\frac{ 21 }{ 2 } \pm \sqrt{(\frac{ 21 }{ 2 })^2 - 98}\]\[x_1 = 7, \space x_2 = 14\]
Now do a quick plot

As you can see in the plot the value cannot be between 7 and 14 beacuse it won't sattisfy \[0 \le x^2 - 21x + 98 \]
x can't be smaler than 2 because sqrt(x-2) won't have soulution in \[\sqrt{x+2} \le x-10\]
Finaly x cannot be smaler than 10 because the right side in is negative which clearly violates the inequillity. \[\sqrt{x+2} \le x-10\]
This excludes all numbrers below 14. Does 14 violate the inequillity?
\[\sqrt{14+2} \le 14-10\]\[4 \le 4\]No!

Hence our answer is \[x \ge 14\]