Problem 50 - Algebra 2

Question

ACT/SAT If \((x+1)(x-2)\) is positive, which statement must be true? A \(x < -1\) or \(x > 2\) B \(x > -1\) or \(x < 2\) C \(-1 < x < 2\) D \(-2 < x < 1\)

Step-by-Step Solution

Answer

Choice A: \(x < -1\) or \(x > 2\).

Step 1: Identify the critical points

The expression given is \((x+1)(x-2)\). To find when this is positive, first figure out where it equals zero. The expression equals zero when either \(x+1=0\) or \(x-2=0\). Solving these gives us critical points at \(x=-1\) and \(x=2\).

Step 2: Determine the intervals

The critical points divide the number line into three intervals: \((-\infty, -1)\), \((-1, 2)\), and \((2, \infty)\). Check each interval to see where the product \((x+1)(x-2)\) is positive.

Step 3: Test each interval for positivity

Choose a test point from each interval: For \((-\infty, -1)\), choose \(x = -2\); for \((-1, 2)\), choose \(x = 0\); and for \((2, \infty)\), choose \(x = 3\). - For \(x = -2\), \((-2+1)(-2-2) = -1 \times -4 = 4\), which is positive.- For \(x = 0\), \((0+1)(0-2) = 1 \times -2 = -2\), which is negative.- For \(x = 3\), \((3+1)(3-2) = 4 \times 1 = 4\), which is positive.

Step 4: Identify the solution intervals

We determined that the product \((x+1)(x-2)\) is positive for \(x < -1\) and \(x > 2\). Thus, the solution is the intervals \((-\infty, -1) \,\cup\, (2, \infty)\).

Step 5: Match the solution to the answer choices

Review the options:- A: \(x < -1\) or \(x > 2\)- B: \(x > -1\) or \(x < 2\)- C: \(-1 < x < 2\)- D: \(-2 < x < 1\)The correct choice is **A**, as it matches the solution intervals.

Key Concepts

Quadratic Expressions

Quadratic expressions are polynomial expressions of degree two, usually in the form of \(ax^2 + bx + c\). In the context of the given exercise, the quadratic expression is expressed in its factored form as \((x+1)(x-2)\). This form makes it easier to find when the expression is zero by setting each factor equal to zero. Understanding the factored form is vital:

• It reveals the roots or zeros of the quadratic, which in the given example are \(-1\) and \(2\).
• It helps in determining the intervals where the expression is positive or negative.

Quadratic expressions can represent a parabola when graphed, and identifying their zero points aids in sketching this curve in terms of direction and intersection with the x-axis. It's crucial to grasp how these expressions can be solved and analyzed, especially when dealing with inequalities.

Critical Points

Critical points in a quadratic inequality are values of \(x\) where the sign of the entire expression could change. For the expression \((x+1)(x-2)\), the critical points are found by setting each factor to zero:

• \(x+1=0\), which gives \(x=-1\)
• \(x-2=0\), which provides \(x=2\)

These points are crucial as they divide the number line into segments, forming boundaries where the expression might change its sign. By identifying critical points, you can effectively apply number line analysis to observe where these sign changes happen and thus solve the inequality.

Number Line Analysis

Number line analysis is a method used to determine the sign or behavior of a mathematical expression across different intervals divided by critical points. For our quadratic expression \((x+1)(x-2)\), our critical points \(x = -1\) and \(x = 2\) divide the number line into three intervals:

• \((\infty, -1)\)
• \((-1, 2)\)
• \((2, \infty)\)

To determine where the expression is positive, we select test points within each interval. For instance:

• In \(\infty, -1)\), selecting \(x = -2\) gives a positive result.
• In \((-1, 2)\), selecting \(x = 0\) results in a negative outcome.
• In \((2, \infty)\), choosing \(x = 3\) gives a positive result.

Through this analysis, we confirm that the expression is positive in the intervals \((\infty, -1)\) and \((2, \infty)\). Number line analysis makes understanding the behavior of the expression over each segment clear and efficient.

Solution Intervals

Solution intervals are where the expression satisfies the inequality. In our example, we determined through number line analysis that the expression \((x+1)(x-2)\) is positive in the intervals \((\infty, -1)\) and \((2, \infty)\). Thus, these intervals are our solution.

• The segment \((\infty, -1)\) indicates all the values less than \(-1\) where the inequality holds true.
• The segment \((2, \infty)\) indicates all values greater than \(2\) where the expression remains positive.

These intervals are essential as they tell us where the inequality statement is true for \(x\). By expressing these intervals in union, we show the complete set of solutions that meet the criteria of the original inequality. Understanding and identifying solution intervals is critical in solving inequalities efficiently.