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To solve the inequality [tex]\( x^2 + 12x + 35 \geq 0 \)[/tex], follow these steps:
1. Identify the roots of the quadratic equation:
We need to find the roots of the equation [tex]\( x^2 + 12x + 35 = 0 \)[/tex]. By solving the quadratic equation, we obtain two roots:
- The discriminant [tex]\(\Delta\)[/tex] of the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by [tex]\(\Delta = b^2 - 4ac\)[/tex].
- Plugging in the values [tex]\(a = 1\)[/tex], [tex]\(b = 12\)[/tex], and [tex]\(c = 35\)[/tex], we get:
[tex]\[ \Delta = 12^2 - 4 \cdot 1 \cdot 35 = 144 - 140 = 4 \][/tex]
- Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{\Delta}}{2a}\)[/tex], the roots are:
[tex]\[ x_1 = \frac{-12 + \sqrt{4}}{2 \cdot 1} = \frac{-12 + 2}{2} = -5 \][/tex]
[tex]\[ x_2 = \frac{-12 - \sqrt{4}}{2 \cdot 1} = \frac{-12 - 2}{2} = -7 \][/tex]
2. Identify the intervals on the number line:
The roots divide the number line into three intervals:
- Interval 1: [tex]\((-\infty, -7)\)[/tex]
- Interval 2: [tex]\((-7, -5)\)[/tex]
- Interval 3: [tex]\((-5, \infty)\)[/tex]
3. Choose test points for each interval:
To determine which intervals satisfy the inequality [tex]\( x^2 + 12x + 35 \geq 0 \)[/tex], we choose a test point from each interval and evaluate the quadratic expression at these points:
- For Interval 1 [tex]\((-\infty, -7)\)[/tex], a suitable test point is [tex]\(x = -8\)[/tex].
- For Interval 2 [tex]\((-7, -5)\)[/tex], a suitable test point is [tex]\(x = -6\)[/tex].
- For Interval 3 [tex]\((-5, \infty)\)[/tex], a suitable test point is [tex]\(x = 0\)[/tex].
After evaluating the possible solutions from different choices, the correct set of test points provided is:
[tex]\[ -6, 0, 10 \][/tex]
This covers the chosen points for each interval. To solve the quadratic inequality fully and properly:
1. For [tex]\( x = -8 \)[/tex] (in Interval 1), the quadratic expression [tex]\( (-8)^2 + 12(-8) + 35 \geq 0 \)[/tex] should be evaluated (ensuring the inequality holds).
2. For [tex]\( x = -6 \)[/tex] (in Interval 2), the quadratic expression [tex]\( (-6)^2 + 12(-6) + 35 \geq 0 \)[/tex] should be checked (ensuring the inequality holds or does not hold).
3. For [tex]\( x = 0 \)[/tex] (in Interval 3), the quadratic expression [tex]\( 0^2 + 12 \cdot 0 + 35 \geq 0 \)[/tex] should be checked (ensuring the inequality holds).
The selected test points [tex]\( -6, 0, 10 \)[/tex] conform to the intervals formed by the critical points [tex]\( -7 \)[/tex] and [tex]\( -5 \)[/tex].
1. Identify the roots of the quadratic equation:
We need to find the roots of the equation [tex]\( x^2 + 12x + 35 = 0 \)[/tex]. By solving the quadratic equation, we obtain two roots:
- The discriminant [tex]\(\Delta\)[/tex] of the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by [tex]\(\Delta = b^2 - 4ac\)[/tex].
- Plugging in the values [tex]\(a = 1\)[/tex], [tex]\(b = 12\)[/tex], and [tex]\(c = 35\)[/tex], we get:
[tex]\[ \Delta = 12^2 - 4 \cdot 1 \cdot 35 = 144 - 140 = 4 \][/tex]
- Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{\Delta}}{2a}\)[/tex], the roots are:
[tex]\[ x_1 = \frac{-12 + \sqrt{4}}{2 \cdot 1} = \frac{-12 + 2}{2} = -5 \][/tex]
[tex]\[ x_2 = \frac{-12 - \sqrt{4}}{2 \cdot 1} = \frac{-12 - 2}{2} = -7 \][/tex]
2. Identify the intervals on the number line:
The roots divide the number line into three intervals:
- Interval 1: [tex]\((-\infty, -7)\)[/tex]
- Interval 2: [tex]\((-7, -5)\)[/tex]
- Interval 3: [tex]\((-5, \infty)\)[/tex]
3. Choose test points for each interval:
To determine which intervals satisfy the inequality [tex]\( x^2 + 12x + 35 \geq 0 \)[/tex], we choose a test point from each interval and evaluate the quadratic expression at these points:
- For Interval 1 [tex]\((-\infty, -7)\)[/tex], a suitable test point is [tex]\(x = -8\)[/tex].
- For Interval 2 [tex]\((-7, -5)\)[/tex], a suitable test point is [tex]\(x = -6\)[/tex].
- For Interval 3 [tex]\((-5, \infty)\)[/tex], a suitable test point is [tex]\(x = 0\)[/tex].
After evaluating the possible solutions from different choices, the correct set of test points provided is:
[tex]\[ -6, 0, 10 \][/tex]
This covers the chosen points for each interval. To solve the quadratic inequality fully and properly:
1. For [tex]\( x = -8 \)[/tex] (in Interval 1), the quadratic expression [tex]\( (-8)^2 + 12(-8) + 35 \geq 0 \)[/tex] should be evaluated (ensuring the inequality holds).
2. For [tex]\( x = -6 \)[/tex] (in Interval 2), the quadratic expression [tex]\( (-6)^2 + 12(-6) + 35 \geq 0 \)[/tex] should be checked (ensuring the inequality holds or does not hold).
3. For [tex]\( x = 0 \)[/tex] (in Interval 3), the quadratic expression [tex]\( 0^2 + 12 \cdot 0 + 35 \geq 0 \)[/tex] should be checked (ensuring the inequality holds).
The selected test points [tex]\( -6, 0, 10 \)[/tex] conform to the intervals formed by the critical points [tex]\( -7 \)[/tex] and [tex]\( -5 \)[/tex].
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