Find The Solution set of X²-2X-8≥0

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∻Solution by Steps∻ ‖ step 1 ⋮ We start with the inequality $ X^2 - 2X - 8 \geq 0 $. ‖ ‖ step 2 ⋮ To solve this, we first find the roots of the equation $ X^2 - 2X - 8 = 0 $ using the quadratic formula $ X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a = 1, b = -2, c = -8 $. ‖ ‖ step 3 ⋮ Calculating the discriminant: $ b^2 - 4ac = (-2)^2 - 4(1)(-8) = 4 + 32 = 36 $. Thus, the roots are $ X = \frac{2 \pm 6}{2} $. ‖ ‖ step 4 ⋮ This gives us the roots $ X = 4 $ and $ X = -2 $. ‖ ‖ step 5 ⋮ We can now test intervals around the roots: $ (-\infty, -2) $, $ (-2, 4) $, and $ (4, \infty) $ to determine where the inequality holds. ‖ ‖ step 6 ⋮ Testing $ X = -3 $ in $ (-\infty, -2) $: $ (-3)^2 - 2(-3) - 8 = 9 + 6 - 8 = 7 \geq 0 $ (true). ‖ ‖ step 7 ⋮ Testing $ X = 0 $ in $ (-2, 4) $: $ 0^2 - 2(0) - 8 = -8 \geq 0 $ (false). ‖ ‖ step 8 ⋮ Testing $ X = 5 $ in $ (4, \infty) $: $ 5^2 - 2(5) - 8 = 25 - 10 - 8 = 7 \geq 0 $ (true). ‖ ‖ step 9 ⋮ Therefore, the solution set is $ X \leq -2 $ or $ X \geq 4 $. ‖ ∻Answer∻ ⚹ The solution set is $ (-\infty, -2] \cup [4, \infty) $ ⚹ ∻Key Concept∻ ⚹ Solving quadratic inequalities involves finding the roots and testing intervals. ⚹ ∻Explanation∻ ⚹ The solution set includes values of $ X $ that satisfy the inequality $ X^2 - 2X - 8 \geq 0 $. The roots divide the number line into intervals, which we test to find where the inequality holds true. ⚹◊What are the critical points of the equation $x^2 - 2x - 8 = 0$?⍭ Generate me a similar question◊

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