Question

Math

Posted 2 months ago

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

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 ≥).

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