Checkpoint Quiz 2.4 Name: 1. Solve: 1 - X² ≤ |x + 2| ✓ Solved

Checkpoint Quiz 2.4 Name: 1. Solve: 1 − x 2 ≤ |x + 2|

1. Solve: 1 - x^2 ≤ |x + 2|. Check your answer graphically.

2. Solve: 16x^2 ≤ 24x - 9.

3. Solve: x^2 > 2(x + 1).

4. Solve: x^2 - 6 ≤ x + |x - 3|.

5. Sketch the relation: R = { (x, y) : x^2 ≤ y

Paper For Above Instructions

Mathematics is a profound and intricate discipline that often poses challenges in both understanding and application. In this paper, we will explore solutions to a variety of inequalities and graphical representations as outlined in the prompt.

1. Inequality: 1 - x² ≤ |x + 2|

To solve the inequality 1 - x² ≤ |x + 2|, we start by breaking the absolute value into two cases.

Case 1: x + 2 ≥ 0 (i.e., x ≥ -2)

In this case, |x + 2| = x + 2. The inequality becomes:

1 - x² ≤ x + 2

Rearranging this gives:

-x² - x - 1 ≤ 0, or x² + x + 1 ≥ 0 (which is always true).

Case 2: x + 2

In this case, |x + 2| = -(x + 2). The inequality becomes:

1 - x² ≤ -x - 2

Rearranging yields:

1 + x + 2 - x² ≤ 0, or x² - x - 3 ≥ 0.

Solve the quadratic equation x² - x - 3 = 0 using the quadratic formula x = [1 ± √(1 + 12)] / 2 = [1 ± √13] / 2. The approximate roots are x ≈ 2 and x ≈ -1.5.

The solution to the inequality is that x ≤ -2 or x ≥ (1 + √13)/2.

Graphically, this can be checked by drawing both sides of the inequality and observing their intersections.

2. Inequality: 16x² ≤ 24x - 9

Next, we solve the inequality 16x² ≤ 24x - 9.

Rearranging gives:

16x² - 24x + 9 ≤ 0.

Using the quadratic formula again:

x = [24 ± √(24² - 4×16×9)] / (2×16) = [24 ± √(576 - 576)] / 32 = [24 ± 0] / 32 = 3/4.

Since this is a perfect square, the solution is simply x = 3/4. Therefore, for this inequality, the solution holds wherever 16(x - 3/4)² ≤ 0, which only occurs at x = 3/4.

3. Inequality: x² > 2(x + 1)

Thus, we solve x² > 2(x + 1) by first rewriting the inequality:

x² - 2x - 2 > 0.

Finding the roots of x² - 2x - 2 = 0 gives:

x = [2 ± √(4 + 8)] / 2 = 1 ± √3.

The roots are approximately x ≈ 3.73 and x ≈ -1.73. The solution set for the inequality is x 3.73, since the parabola opens upwards.

4. Inequality: x² - 6 ≤ x + |x - 3|

The inequality x² - 6 ≤ x + |x - 3| also needs to be broken down into cases.

Case 1: x - 3 ≥ 0 (i.e., x ≥ 3): In this case, |x - 3| = x - 3. The inequality becomes:

x² - 6 ≤ x + x - 3, or x² - 2x - 3 ≤ 0.

Factoring yields (x - 3)(x + 1) ≤ 0, giving us the intervals x ∈ [-1, 3].

Case 2: x - 3

x² - 6 ≤ x - x + 3, or x² ≤ 9.

This simplifies to -3 ≤ x ≤ 3, thus providing another valid interval.

5. Relation Sketch: R = { (x, y) : x² ≤ y

For this relation R, we need to sketch the inequalities x² ≤ y and y

The first inequality represents the area above the parabola y = x²; the second inequality represents the area below the line y = x + 6. The area that satisfies both inequalities will be between these curves.

Conclusion

In conclusion, through solving and graphically analyzing these mathematical inequalities, we have reinforced crucial skills in understanding complex relationships between variables. It has been proven that graphical solutions can greatly enhance comprehension in mathematical reasoning.

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