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To solve the inequality [tex]\(x^2 - 2x - 3 > 0\)[/tex], we will need to determine the intervals where the quadratic expression is greater than zero. Let's go through the process step-by-step:
1. Factorize the quadratic expression:
Start by factoring the quadratic expression [tex]\(x^2 - 2x - 3\)[/tex].
We need to find two numbers that multiply to [tex]\(-3\)[/tex] (the constant term) and add up to [tex]\(-2\)[/tex] (the coefficient of the [tex]\(x\)[/tex] term).
These numbers are [tex]\(-3\)[/tex] and [tex]\(1\)[/tex], since [tex]\((-3) \cdot 1 = -3\)[/tex] and [tex]\((-3) + 1 = -2\)[/tex].
So, we can write:
[tex]\[ x^2 - 2x - 3 = (x - 3)(x + 1) \][/tex]
2. Set up the inequality:
Now, we need to find when the product of these factors is greater than zero:
[tex]\[ (x - 3)(x + 1) > 0 \][/tex]
3. Determine the critical points:
The critical points are the values of [tex]\(x\)[/tex] that make the expression equal to zero:
[tex]\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \][/tex]
[tex]\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \][/tex]
4. Analyze the intervals:
The critical points divide the number line into three intervals:
- [tex]\(x < -1\)[/tex]
- [tex]\(-1 < x < 3\)[/tex]
- [tex]\(x > 3\)[/tex]
We will test each of these intervals to see where the product [tex]\((x - 3)(x + 1)\)[/tex] is positive.
- For [tex]\(x < -1\)[/tex], choose a test point like [tex]\(x = -2\)[/tex]:
[tex]\[ (x - 3)(x + 1) \quad \text{becomes} \quad (-2 - 3)(-2 + 1) = (-5)(-1) > 0 \][/tex]
- For [tex]\(-1 < x < 3\)[/tex], choose a test point like [tex]\(x = 0\)[/tex]:
[tex]\[ (x - 3)(x + 1) \quad \text{becomes} \quad (0 - 3)(0 + 1) = (-3)(1) < 0 \][/tex]
- For [tex]\(x > 3\)[/tex], choose a test point like [tex]\(x = 4\)[/tex]:
[tex]\[ (x - 3)(x + 1) \quad \text{becomes} \quad (4 - 3)(4 + 1) = (1)(5) > 0 \][/tex]
5. Combine the intervals:
From our tests, we see that the expression [tex]\((x - 3)(x + 1)\)[/tex] is positive for:
[tex]\[ x < -1 \quad \text{or} \quad x > 3 \][/tex]
Therefore, the solution to the inequality [tex]\(x^2 - 2x - 3 > 0\)[/tex] is:
[tex]\[ x < -1 \quad \text{or} \quad x > 3 \][/tex]
1. Factorize the quadratic expression:
Start by factoring the quadratic expression [tex]\(x^2 - 2x - 3\)[/tex].
We need to find two numbers that multiply to [tex]\(-3\)[/tex] (the constant term) and add up to [tex]\(-2\)[/tex] (the coefficient of the [tex]\(x\)[/tex] term).
These numbers are [tex]\(-3\)[/tex] and [tex]\(1\)[/tex], since [tex]\((-3) \cdot 1 = -3\)[/tex] and [tex]\((-3) + 1 = -2\)[/tex].
So, we can write:
[tex]\[ x^2 - 2x - 3 = (x - 3)(x + 1) \][/tex]
2. Set up the inequality:
Now, we need to find when the product of these factors is greater than zero:
[tex]\[ (x - 3)(x + 1) > 0 \][/tex]
3. Determine the critical points:
The critical points are the values of [tex]\(x\)[/tex] that make the expression equal to zero:
[tex]\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \][/tex]
[tex]\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \][/tex]
4. Analyze the intervals:
The critical points divide the number line into three intervals:
- [tex]\(x < -1\)[/tex]
- [tex]\(-1 < x < 3\)[/tex]
- [tex]\(x > 3\)[/tex]
We will test each of these intervals to see where the product [tex]\((x - 3)(x + 1)\)[/tex] is positive.
- For [tex]\(x < -1\)[/tex], choose a test point like [tex]\(x = -2\)[/tex]:
[tex]\[ (x - 3)(x + 1) \quad \text{becomes} \quad (-2 - 3)(-2 + 1) = (-5)(-1) > 0 \][/tex]
- For [tex]\(-1 < x < 3\)[/tex], choose a test point like [tex]\(x = 0\)[/tex]:
[tex]\[ (x - 3)(x + 1) \quad \text{becomes} \quad (0 - 3)(0 + 1) = (-3)(1) < 0 \][/tex]
- For [tex]\(x > 3\)[/tex], choose a test point like [tex]\(x = 4\)[/tex]:
[tex]\[ (x - 3)(x + 1) \quad \text{becomes} \quad (4 - 3)(4 + 1) = (1)(5) > 0 \][/tex]
5. Combine the intervals:
From our tests, we see that the expression [tex]\((x - 3)(x + 1)\)[/tex] is positive for:
[tex]\[ x < -1 \quad \text{or} \quad x > 3 \][/tex]
Therefore, the solution to the inequality [tex]\(x^2 - 2x - 3 > 0\)[/tex] is:
[tex]\[ x < -1 \quad \text{or} \quad x > 3 \][/tex]
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