Problem 2 - Algebra and Trigonometry

Question

The solutions of the inequality \(x^{2}-2 x-3>0\) are the \(x\) -coordinates of the points on the graph of \(y=x^{2}-2 x-3\) that lie ______the \(x\) -axis.

Step-by-Step Solution

Answer

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Step 1: Identify the problem type

We are given the inequality \(x^{2}-2x-3>0\). This is a quadratic inequality, which means we need to find the range of \(x\) values where this inequality holds true.

Step 2: Factor the quadratic expression

The quadratic expression \(x^{2}-2x-3\) can be factored to \((x-3)(x+1)\). We rewrite the inequality as \((x-3)(x+1)>0\).

Step 3: Determine the critical points

The critical points are the solutions of the equation \((x-3)(x+1)=0\). Thus, the critical points are \(x=3\) and \(x=-1\).

Step 4: Test intervals between critical points

The critical points divide the number line into three intervals: \((-\infty, -1)\), \((-1, 3)\), and \((3, \infty)\). We test these intervals to determine where the inequality holds.- Choose a test point from each interval. - For \((-\infty, -1)\), choose \(x = -2\). Plug into the factored form: \((-2-3)(-2+1) = 5 > 0\). - For \((-1, 3)\), choose \(x = 0\). Plug into the factored form: \((0-3)(0+1) = -3 < 0\). - For \((3, \infty)\), choose \(x = 4\). Plug into the factored form: \((4-3)(4+1) = 5 > 0\).

Step 5: Determine solution set from tested intervals

The inequality \((x-3)(x+1)>0\) holds in the intervals \((-\infty, -1)\) and \((3, \infty)\) where the test points satisfied the inequality.

Step 6: Relate solution to the graph of the function

The values of \(x\) that solve the inequality correspond to the parts of the graph of \(y = x^{2} - 2x - 3\) that lie above the \(x\)-axis. This is because the inequality is \(>0\), indicating regions where \(y\) is positive.

Key Concepts

Factoring Quadratic Expressions

Factoring quadratic expressions is crucial for solving many quadratic equations and inequalities. In our exercise, we begin with the quadratic expression \(x^2 - 2x - 3\). To factor it, we need to find two numbers that multiply to the constant term (\(-3\)) and add up to the linear coefficient (\(-2\)).
This gives us the factors \((-3)\) and \(1\), as \((-3) + 1 = -2\). Thus, the quadratic expression can be rewritten as \((x-3)(x+1)\).
Factoring is like finding the building blocks of our quadratic expression, which makes it easier to analyze and solve the related inequality. It provides a way to see when the quadratic expression becomes zero, which is essential for determining the critical points in the next step.

Critical Points

Critical points play a key role in understanding quadratic inequalities, as they help us determine "boundaries" on the number line. These are the values of \(x\) at which the quadratic expression equals zero.
To find the critical points, set \((x-3)(x+1) = 0\). This equation implies either \(x-3=0\) or \(x+1=0\). Solving these gives \(x=3\) and \(x=-1\).
The critical points, \(x=3\) and \(x=-1\), are where our quadratic may change sign. This means they help us identify the intervals of interest on the number line. These points act like landmarks that split the real number line into distinct regions, which we will analyze in the next step to understand where the inequality holds.

Testing Intervals

Testing intervals is the final step in solving a quadratic inequality. It involves checking if the inequality holds in each of the intervals created by the critical points.
Our critical points, \(x=3\) and \(x=-1\), divide the number line into three intervals: \((-fty, -1)\), \((-1, 3)\), and \((3, fty)\). For each interval, we choose a test point and plug it back into the factored form \((x-3)(x+1)\) to see if it satisfies the inequality \((x-3)(x+1) > 0\).

• Pick \(x = -2\) for the interval \((-fty, -1)\), and calculate: \((-2-3)(-2+1) = 5 > 0\). The inequality holds.
• Pick \(x = 0\) for the interval \((-1, 3)\), and calculate: \((0-3)(0+1) = -3 < 0\). The inequality does not hold.
• Pick \(x = 4\) for the interval \((3, fty)\), and calculate: \((4-3)(4+1) = 5 > 0\). The inequality holds.

By testing these intervals, we find that the inequality holds for \((-fty, -1)\) and \((3, fty)\). These are the solution sets, representing the \(x\)-values where the original expression is greater than zero and where the graph of the function \(y = x^2 - 2x - 3\) is above the \(x\)-axis.