The solution set of the inequation sqrt{x^2+6 | vedclass

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(C) For the square root to be defined,$x^2+6x+5 \ge 0$,which implies $(x+5)(x+1) \ge 0$,so $x \in (-\infty, -5] \cup [-1, \infty)$.For the inequality

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$(\frac{59}{22}, \infty)$

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(C) For the square root to be defined,$x^2+6x+5 \ge 0$,which implies $(x+5)(x+1) \ge 0$,so $x \in (-\infty, -5] \cup [-1, \infty)$.For the inequality $\sqrt{x^2+6x+5} > (8-x)$ to hold,we must have $8-x  (8-x)^2$).Case $1$: $8-x  8$. Since $x > 8$ satisfies $x \in [-1, \infty)$,this is a valid solution.Case $2$: $8-x \ge 0 \implies x \le 8$. Squaring both sides: $x^2+6x+5 > 64-16x+x^2$.$22x > 59 \implies x > \frac{59}{22}$.Combining $x > \frac{59}{22}$ with $x \le 8$,we get $x \in (\frac{59}{22}, 8]$.Combining Case $1$ and Case $2$: $x \in (\frac{59}{22}, 8] \cup (8, \infty) = (\frac{59}{22}, \infty)$.

The solution set of the inequation $\sqrt{x^2+6x+5} > (8-x)$ is

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