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Sagot :
To find the values of [tex]\( x \)[/tex] that satisfy both inequalities:
[tex]\[ \begin{array}{l} x^2-2 x-24<0 \\ -x^2+9 x-14 \geq 0 \end{array} \][/tex]
we will solve each inequality separately and then find their intersection.
### Step 1: Solve the First Inequality
The first inequality is:
[tex]\[ x^2 - 2x - 24 < 0 \][/tex]
1. Find the roots of the equation [tex]\( x^2 - 2x - 24 = 0 \)[/tex]:
[tex]\[ x^2 - 2x - 24 = 0 \][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -24 \)[/tex]:
[tex]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-24)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{4 + 96}}{2} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{100}}{2} \][/tex]
[tex]\[ x = \frac{2 \pm 10}{2} \][/tex]
[tex]\[ x = \frac{12}{2} \quad \text{or} \quad x = \frac{-8}{2} \][/tex]
[tex]\[ x = 6 \quad \text{or} \quad x = -4 \][/tex]
2. Determine the intervals for the inequality [tex]\( x^2 - 2x - 24 < 0 \)[/tex]:
The roots divide the number line into three intervals: [tex]\( (-\infty, -4) \)[/tex], [tex]\( (-4, 6) \)[/tex], and [tex]\( (6, \infty) \)[/tex]. We test a point in each interval to determine where the inequality holds.
- For [tex]\( x < -4 \)[/tex] (e.g., [tex]\( x = -5 \)[/tex]):
[tex]\[ (-5)^2 - 2(-5) - 24 = 25 + 10 - 24 = 11 > 0 \][/tex]
- For [tex]\( -4 < x < 6 \)[/tex] (e.g., [tex]\( x = 0 \)[/tex]):
[tex]\[ 0^2 - 2(0) - 24 = -24 < 0 \][/tex]
- For [tex]\( x > 6 \)[/tex] (e.g., [tex]\( x = 7 \)[/tex]):
[tex]\[ 7^2 - 2(7) - 24 = 49 - 14 - 24 = 11 > 0 \][/tex]
Therefore, the solution for the first inequality is:
[tex]\[ -4 < x < 6 \][/tex]
### Step 2: Solve the Second Inequality
The second inequality is:
[tex]\[ -x^2 + 9x - 14 \geq 0 \][/tex]
1. Find the roots of the equation [tex]\( -x^2 + 9x - 14 = 0 \)[/tex]:
[tex]\[ -x^2 + 9x - 14 = 0 \][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = -1 \)[/tex], [tex]\( b = 9 \)[/tex], and [tex]\( c = -14 \)[/tex]:
[tex]\[ x = \frac{-9 \pm \sqrt{9^2 - 4(-1)(-14)}}{2(-1)} \][/tex]
[tex]\[ x = \frac{-9 \pm \sqrt{81 - 56}}{-2} \][/tex]
[tex]\[ x = \frac{-9 \pm \sqrt{25}}{-2} \][/tex]
[tex]\[ x = \frac{-9 \pm 5}{-2} \][/tex]
[tex]\[ x = \frac{-9 + 5}{-2} \quad \text{or} \quad x = \frac{-9 - 5}{-2} \][/tex]
[tex]\[ x = \frac{-4}{-2} \quad \text{or} \quad x = \frac{-14}{-2} \][/tex]
[tex]\[ x = 2 \quad \text{or} \quad x = 7 \][/tex]
2. Determine the intervals for the inequality [tex]\( -x^2 + 9x - 14 \geq 0 \)[/tex]:
The roots divide the number line into three intervals: [tex]\( (-\infty, 2) \)[/tex], [tex]\( (2, 7) \)[/tex], and [tex]\( (7, \infty) \)[/tex]. We test a point in each interval to determine where the inequality holds.
- For [tex]\( x < 2 \)[/tex] (e.g., [tex]\( x = 1 \)[/tex]):
[tex]\[ -(1)^2 + 9(1) - 14 = -1 + 9 - 14 = -6 < 0 \][/tex]
- For [tex]\( 2 < x < 7 \)[/tex] (e.g., [tex]\( x = 3 \)[/tex]):
[tex]\[ -(3)^2 + 9(3) - 14 = -9 + 27 - 14 = 4 \geq 0 \][/tex]
- For [tex]\( x > 7 \)[/tex] (e.g., [tex]\( x = 8 \)[/tex]):
[tex]\[ -(8)^2 + 9(8) - 14 = -64 + 72 - 14 = -6 < 0 \][/tex]
Therefore, the solution for the second inequality is:
[tex]\[ 2 \leq x \leq 7 \][/tex]
### Step 3: Find the Intersection of the Solutions
- First inequality: [tex]\( -4 < x < 6 \)[/tex]
- Second inequality: [tex]\( 2 \leq x \leq 7 \)[/tex]
The intersection of these solutions is:
[tex]\[ 2 \leq x < 6 \][/tex]
### Final Answer
The values of [tex]\( x \)[/tex] which satisfy both inequalities are:
[tex]\[ 2 \leq x < 6 \][/tex]
[tex]\[ \begin{array}{l} x^2-2 x-24<0 \\ -x^2+9 x-14 \geq 0 \end{array} \][/tex]
we will solve each inequality separately and then find their intersection.
### Step 1: Solve the First Inequality
The first inequality is:
[tex]\[ x^2 - 2x - 24 < 0 \][/tex]
1. Find the roots of the equation [tex]\( x^2 - 2x - 24 = 0 \)[/tex]:
[tex]\[ x^2 - 2x - 24 = 0 \][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -24 \)[/tex]:
[tex]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-24)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{4 + 96}}{2} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{100}}{2} \][/tex]
[tex]\[ x = \frac{2 \pm 10}{2} \][/tex]
[tex]\[ x = \frac{12}{2} \quad \text{or} \quad x = \frac{-8}{2} \][/tex]
[tex]\[ x = 6 \quad \text{or} \quad x = -4 \][/tex]
2. Determine the intervals for the inequality [tex]\( x^2 - 2x - 24 < 0 \)[/tex]:
The roots divide the number line into three intervals: [tex]\( (-\infty, -4) \)[/tex], [tex]\( (-4, 6) \)[/tex], and [tex]\( (6, \infty) \)[/tex]. We test a point in each interval to determine where the inequality holds.
- For [tex]\( x < -4 \)[/tex] (e.g., [tex]\( x = -5 \)[/tex]):
[tex]\[ (-5)^2 - 2(-5) - 24 = 25 + 10 - 24 = 11 > 0 \][/tex]
- For [tex]\( -4 < x < 6 \)[/tex] (e.g., [tex]\( x = 0 \)[/tex]):
[tex]\[ 0^2 - 2(0) - 24 = -24 < 0 \][/tex]
- For [tex]\( x > 6 \)[/tex] (e.g., [tex]\( x = 7 \)[/tex]):
[tex]\[ 7^2 - 2(7) - 24 = 49 - 14 - 24 = 11 > 0 \][/tex]
Therefore, the solution for the first inequality is:
[tex]\[ -4 < x < 6 \][/tex]
### Step 2: Solve the Second Inequality
The second inequality is:
[tex]\[ -x^2 + 9x - 14 \geq 0 \][/tex]
1. Find the roots of the equation [tex]\( -x^2 + 9x - 14 = 0 \)[/tex]:
[tex]\[ -x^2 + 9x - 14 = 0 \][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = -1 \)[/tex], [tex]\( b = 9 \)[/tex], and [tex]\( c = -14 \)[/tex]:
[tex]\[ x = \frac{-9 \pm \sqrt{9^2 - 4(-1)(-14)}}{2(-1)} \][/tex]
[tex]\[ x = \frac{-9 \pm \sqrt{81 - 56}}{-2} \][/tex]
[tex]\[ x = \frac{-9 \pm \sqrt{25}}{-2} \][/tex]
[tex]\[ x = \frac{-9 \pm 5}{-2} \][/tex]
[tex]\[ x = \frac{-9 + 5}{-2} \quad \text{or} \quad x = \frac{-9 - 5}{-2} \][/tex]
[tex]\[ x = \frac{-4}{-2} \quad \text{or} \quad x = \frac{-14}{-2} \][/tex]
[tex]\[ x = 2 \quad \text{or} \quad x = 7 \][/tex]
2. Determine the intervals for the inequality [tex]\( -x^2 + 9x - 14 \geq 0 \)[/tex]:
The roots divide the number line into three intervals: [tex]\( (-\infty, 2) \)[/tex], [tex]\( (2, 7) \)[/tex], and [tex]\( (7, \infty) \)[/tex]. We test a point in each interval to determine where the inequality holds.
- For [tex]\( x < 2 \)[/tex] (e.g., [tex]\( x = 1 \)[/tex]):
[tex]\[ -(1)^2 + 9(1) - 14 = -1 + 9 - 14 = -6 < 0 \][/tex]
- For [tex]\( 2 < x < 7 \)[/tex] (e.g., [tex]\( x = 3 \)[/tex]):
[tex]\[ -(3)^2 + 9(3) - 14 = -9 + 27 - 14 = 4 \geq 0 \][/tex]
- For [tex]\( x > 7 \)[/tex] (e.g., [tex]\( x = 8 \)[/tex]):
[tex]\[ -(8)^2 + 9(8) - 14 = -64 + 72 - 14 = -6 < 0 \][/tex]
Therefore, the solution for the second inequality is:
[tex]\[ 2 \leq x \leq 7 \][/tex]
### Step 3: Find the Intersection of the Solutions
- First inequality: [tex]\( -4 < x < 6 \)[/tex]
- Second inequality: [tex]\( 2 \leq x \leq 7 \)[/tex]
The intersection of these solutions is:
[tex]\[ 2 \leq x < 6 \][/tex]
### Final Answer
The values of [tex]\( x \)[/tex] which satisfy both inequalities are:
[tex]\[ 2 \leq x < 6 \][/tex]
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