Questions: Solve (2 x^2+16 x-1>-2 x^2+12 x+47). Enter the solution in interval notation.

Solution Steps

To solve the inequality \(2x^2 + 16x - 1 > -2x^2 + 12x + 47\), we first move all terms to one side to form a single quadratic inequality. Then, we solve the corresponding quadratic equation to find the critical points. These points divide the number line into intervals, and we test each interval to determine where the inequality holds true. Finally, we express the solution in interval notation.

We start with the inequality: \[ 2x^2 + 16x - 1 > -2x^2 + 12x + 47 \] Rearranging all terms to one side gives: \[ 2x^2 + 16x - 1 + 2x^2 - 12x - 47 > 0 \] This simplifies to: \[ 4x^2 + 4x - 48 > 0 \] Dividing the entire inequality by 4 results in: \[ x^2 + x - 12 > 0 \]

Next, we solve the corresponding equation: \[ x^2 + x - 12 = 0 \] Factoring gives: \[ (x - 3)(x + 4) = 0 \] Thus, the critical points are: \[ x = 3 \quad \text{and} \quad x = -4 \]

The critical points divide the number line into intervals: \((- \infty, -4)\), \((-4, 3)\), and \((3, \infty)\). We test each interval to determine where the inequality holds.

1. For the interval \((- \infty, -4)\), choose \(x = -5\): \[ (-5)^2 + (-5) - 12 = 25 - 5 - 12 = 8 > 0 \quad \text{(True)} \]

2. For the interval \((-4, 3)\), choose \(x = 0\): \[ 0^2 + 0 - 12 = -12 < 0 \quad \text{(False)} \]

3. For the interval \((3, \infty)\), choose \(x = 4\): \[ 4^2 + 4 - 12 = 16 + 4 - 12 = 8 > 0 \quad \text{(True)} \]

Final Answer