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sqrt(x^(2)+2x-3) >= -2

$\sqrt{x^{2}+2 x-3} \geq-2$

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Add and subtract polynomials

Full solution

Q.

\sqrt{x^{2}+2 x-3} \geq-2

Simplify inequality: Simplify the inequality by noting that the square root function, $\sqrt{}$ , is always non-negative ( $\geq 0$ ). Therefore, the inequality $\sqrt{x^{2}+2x-3} \geq -2$ is always true for all $x$ where the expression under the square root is non-negative.
Solve quadratic expression: Solve the inequality $x^{2}+2x-3 \geq 0$ . This is a quadratic inequality. Factorize the quadratic expression.
Determine critical points: Determine the critical points where $(x+3)(x-1) = 0$ . Solving for $x$ gives $x = -3$ and $x = 1$ .
Test intervals with points: Test the intervals determined by the critical points: $(-\infty, -3)$ , $(-3, 1)$ , and $(1, \infty)$ . Use test points $x = -4$ , $x = 0$ , and $x = 2$ respectively.
Final solution: From the test points, the expression $(x+3)(x-1)$ is positive in the intervals $(-\infty, -3)$ and $(1, \infty)$ . Thus, the solution to the inequality $x^{2}+2x-3 \geq 0$ is $x \in (-\infty, -3] \cup [1, \infty)$ .

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