Solve the following quadratic inequality: $x^2 - x$ - 4 ≤ 2

Analysis

To solve the quadratic inequality $x^2 - x - 4 \leq 2$, we need to move all terms to one side to set the inequality to zero, find the roots of the corresponding quadratic equation, and then determine the interval that satisfies the condition.

Concept

A quadratic inequality in the form $ax^2 + bx + c \leq 0$ is solved by:

1. Finding the roots of the equation $ax^2 + bx + c = 0$.
2. Testing the intervals created by these roots to see which one satisfies the inequality.
3. Since the inequality is "$\leq$", the solution will include the roots themselves (using $\leq$ or $\geq$ signs).

Step-by-Step Solution

Step 1: Rearrange the inequality Subtract 2 from both sides to set the right side to zero: $x^2 - x - 4 - 2 \leq 0$ $x^2 - x - 6 \leq 0$

Step 2: Factor the quadratic expression We need two numbers that multiply to $-6$ and add up to $-1$. These numbers are $-3$ and $+2$. $(x - 3)(x + 2) \leq 0$

Step 3: Find the critical points Set each factor to zero: $x - 3 = 0 \implies x = 3$ $x + 2 = 0 \implies x = -2$

Step 4: Determine the interval We are looking for where the product $(x - 3)(x + 2)$ is less than or equal to zero (negative or zero).

• If $x < -2$ (e.g., $x = -3$): $(-3-3)(-3+2) = (-6)(-1) = 6$ (Positive)
• If $-2 \leq x \leq 3$ (e.g., $x = 0$): $(0-3)(0+2) = (-3)(2) = -6$ (Negative)
• If $x > 3$ (e.g., $x = 4$): $(4-3)(4+2) = (1)(6) = 6$ (Positive)

Since we want the expression to be $\leq 0$, the valid range is between the two roots, inclusive. Thus, $-2 \leq x \leq 3$.

Conclusion

The solution to the inequality is $-2 \leq x \leq 3$. This matches option B.

Exam Tip: Whenever you have a quadratic inequality in the form $(x - a)(x - b) \leq 0$ (where $a < b$), the solution is almost always the interval between the roots: $a \leq x \leq b$. If the sign were $\geq 0$, the solution would be the "outer" regions: $x \leq a$ or $x \geq b$.