Question

The common solution set of the inequalities \[ x^2 - 4x \leq 12 \quad \text{and} \quad x^2 - 2x \geq 15 \] taken together is:

• A. \( (5,6) \)
• B. \( [5,6] \)
• C. \( [-3,5] \)
• D. \( (-\infty,-3] \cup [5,\infty) \)

Hint:

For quadratic inequalities, always factorize and use sign analysis to determine the solution intervals. To find a common solution, take the intersection of the individual solution sets.

Answer 1

The Correct Option is B

We need to find the common solution set of the given inequalities: \[ x^2 - 4x \leq 12 \] \[ x^2 - 2x \geq 15. \]

 Step 1: Solve the First Inequality 
Rearrange: \[ x^2 - 4x - 12 \leq 0. \] Factorizing: \[ (x - 6)(x + 2) \leq 0. \] Using the sign analysis method, the solution is: \[ -2 \leq x \leq 6. \] 

Step 2: Solve the Second Inequality 
Rearrange: \[ x^2 - 2x - 15 \geq 0. \] Factorizing: \[ (x - 5)(x + 3) \geq 0. \] Using the sign analysis method, the solution is: \[ x \leq -3 \quad \text{or} \quad x \geq 5. \] 

Step 3: Find the Common Solution 
The intersection of the two solution sets: \[ -2 \leq x \leq 6 \quad \text{and} \quad x \leq -3 \text{ or } x \geq 5. \] The common region is: \[ 5 \leq x \leq 6. \] Thus, the final solution is: \[ [5,6]. \]