Consider the quadratic inequality $x^{2}-7 x+6 \leq 0$.

What is the solution set for this inequality?

Question

Consider the quadratic inequality $x^{2}-7 x+6 \leq 0$.

What is the solution set for this inequality?

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Consider the quadratic inequality $x^2 - 7x + 6 \leq 0$. ‖ ‖ step 2 ⋮ Factor the quadratic expression: $x^2 - 7x + 6 = (x - 1)(x - 6)$. ‖ ‖ step 3 ⋮ The inequality becomes $(x - 1)(x - 6) \leq 0$. ‖ ‖ step 4 ⋮ Determine the critical points by setting each factor equal to zero: $x - 1 = 0$ and $x - 6 = 0$. Thus, the critical points are $x = 1$ and $x = 6$. ‖ ‖ step 5 ⋮ Test the intervals determined by the critical points: $(-\infty, 1)$, $(1, 6)$, and $(6, \infty)$. ‖ ‖ step 6 ⋮ Choose a test point in each interval and substitute it into the inequality $(x - 1)(x - 6) \leq 0$: - For $x \in (-\infty, 1)$, choose $x = 0$: $(0 - 1)(0 - 6) = 6 > 0$ (not in the solution set). - For $x \in (1, 6)$, choose $x = 3$: $(3 - 1)(3 - 6) = -6 \leq 0$ (in the solution set). - For $x \in (6, \infty)$, choose $x = 7$: $(7 - 1)(7 - 6) = 6 > 0$ (not in the solution set). ‖ ‖ step 7 ⋮ Include the critical points $x = 1$ and $x = 6$ in the solution set because the inequality is $\leq$ (less than or equal to). ‖ ‖ step 8 ⋮ The solution set is $1 \leq x \leq 6$. ‖ ∻Answer∻ ⚹ $1 \leq x \leq 6$ ⚹ ∻Key Concept∻ ⚹Quadratic Inequality⚹ ∻Explanation∻ ⚹To solve a quadratic inequality, factor the quadratic expression, find the critical points, test the intervals, and include the critical points if the inequality is non-strict (≤ or ≥).⚹◊What are the roots of the quadratic equation $x^{2}-7x+6=0$⍭ Generate me a similar question◊

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