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

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(C) For the square root to be defined,we must have $x^2+x-2 \ge 0$,which implies $(x+2)(x-1) \ge 0$. Thus,$x \in (-\infty, -2] \cup [1, \infty)$.Case

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$(1, \infty)$

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(C) For the square root to be defined,we must have $x^2+x-2 \ge 0$,which implies $(x+2)(x-1) \ge 0$. Thus,$x \in (-\infty, -2] \cup [1, \infty)$.Case $1$: If $1-x  1$,the inequality $\sqrt{x^2+x-2} > 1-x$ is always true because the left side is non-negative and the right side is negative.Case $2$: If $1-x \ge 0$,i.e.,$x \le 1$,we square both sides: $x^2+x-2 > (1-x)^2$.$x^2+x-2 > 1-2x+x^2$.$x-2 > 1-2x$ $\Rightarrow 3x > 3$ $\Rightarrow x > 1$.Combining this with the condition $x \le 1$,we get no solution from this case.However,considering the domain $x \in (-\infty, -2] \cup [1, \infty)$,for $x > 1$,the inequality holds.Checking $x \le -2$: If $x = -2$,$\sqrt{4-2-2} = 0$ and $1-(-2) = 3$. $0 > 3$ is false.Thus,the solution set is $(1, \infty)$.

The solution set of the inequation $\sqrt{x^2+x-2} > (1-x)$ is

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