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Question
Math
Posted 2 months ago

Consider the quadratic inequality x27x+60x^{2}-7 x+6 \leq 0.

What is the solution set for this inequality?
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Answer from Sia
Posted 2 months ago
Solution by Steps
step 1
Consider the quadratic inequality x27x+60x^2 - 7x + 6 \leq 0
step 2
Factor the quadratic expression: x27x+6=(x1)(x6)x^2 - 7x + 6 = (x - 1)(x - 6)
step 3
The inequality becomes (x1)(x6)0(x - 1)(x - 6) \leq 0
step 4
Determine the critical points by setting each factor equal to zero: x1=0x - 1 = 0 and x6=0x - 6 = 0. Thus, the critical points are x=1x = 1 and x=6x = 6
step 5
Test the intervals determined by the critical points: (,1)(-\infty, 1), (1,6)(1, 6), and (6,)(6, \infty)
step 6
Choose a test point in each interval and substitute it into the inequality (x1)(x6)0(x - 1)(x - 6) \leq 0: - For x(,1)x \in (-\infty, 1), choose x=0x = 0: (01)(06)=6>0(0 - 1)(0 - 6) = 6 > 0 (not in the solution set). - For x(1,6)x \in (1, 6), choose x=3x = 3: (31)(36)=60(3 - 1)(3 - 6) = -6 \leq 0 (in the solution set). - For x(6,)x \in (6, \infty), choose x=7x = 7: (71)(76)=6>0(7 - 1)(7 - 6) = 6 > 0 (not in the solution set)
step 7
Include the critical points x=1x = 1 and x=6x = 6 in the solution set because the inequality is \leq (less than or equal to)
step 8
The solution set is 1x61 \leq x \leq 6
Answer
1x61 \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 ≥).

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