Quadratic inequality
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This document discusses solving quadratic inequalities. It provides examples of single-variable quadratic inequalities and explains how to find the solution set by first setting the inequality equal to an equation, solving for the roots, and then testing values within the intervals formed by the roots. The document also introduces quadratic inequalities with two variables and how to represent them. It defines a quadratic inequality and explains the process of solving them by relating it back to solving a quadratic equation.
Quadratic inequality
- 2. Activity 1: What Makes Me True? Give the solution/s of each of the following mathematical sentences. 1. x + 5 > 8 2. r – 3 = 10 3. 2s + 7 ≥ 21 4. 3t – 2 ≤ 13 5. 12 – 5m = - 8
- 3. Guide Questions How did you find the solution/s of each mathematical statements? What mathematical concepts or principles did you apply to come up with the solution/s? Which mathematical sentences has only one solution? More than one solution? Describe these sentences.
- 4. Activity 2: Which are Not Quadratic Equations? x2 + 9z + 20 = 0 2r2 < 21 - 9t 2x2 + 2 = 10x r2 + 10r ≤ - 16 m2 = 6m - 7 4x2 – 25 = 0 15 – 6h2 = 10 3w2 + 12w ≥ 0 2s2 + 7s + 5 > 0
- 5. Definition Is an inequality that contains a polynomial of degree 2 and can be written in any of the following forms. ax2 + bx + c > 0 ax2 + bx + c ≥ 0 ax2 + bx + c < 0 ax2 + bx + c ≤ 0 where a, b, and c are real numbers and a ≠ 0.
- 6. To solve a quadratic inequality, find the roots of its corresponding equality. Find the solution set of x2 + 7x + 12 > 0. The corresponding equality of x2 + 7x + 12 > 0 is x2 + 7x + 12 = 0. Solve x2 + 7x + 12 = 0. (x + 3)(x + 4) = 0 Why? x + 3 = 0 & x + 4 = 0 Why? x = - 3 & x = - 4 Why?
- 7. Plot the points corresponding to -3 and -4 on the number line. The three interval are: - ∞ < x < - 4, - 4 < x < - 3, - 3 < x < ∞. Test a number from each interval against the inequality. For - ∞ < x < - 4, Let x = - 7 For – 4 < x < - 3, Let x = 3.6 For – 3 < x < ∞, Let x = 0 x2 + 7x + 12 > 0 (-7)2 + 7(-7) + 12 > 0 49 – 49 + 12 > 0 12 > 0 (true) x2 + 7x + 12 > 0 (-3.6)2 + 7(-3.6) + 12 > 0 12.96 – 25.2 + 12 > 0 -0.24 > 0 (false) x2 + 7x + 12 > 0 (0)2 + 7(0) + 12 > 0 0 + 0 + 12 > 0 12 > 0 (true)
- 8. Also test whether the points x = - 3 and x = - 4 satisfy the equation. x2 + 7x + 12 > 0 (-3)2 + 7(-3) + 12 > 0 9 – 21 + 12 > 0 0 > 0 (false) x2 + 7x + 12 > 0 (-4)2 + 7(-4) + 12 > 0 16 – 28 + 12 > 0 0 > 0 (false) Therefore, the inequality is true for any value of x in the interval - ∞ < x < - 4 or - 3 < x < ∞, and these intervals exclude – 3 and – 4. The solution set of the inequality is {x:x < - 4 or x > - 3}.
- 9. Quadratic Inequalities In Two Variables There are quadratic inequalities that involves two variables. These inequalities can be written in any of the following forms. y > ax2 + bx + c y ≥ ax2 + bx + c y < ax2 + bx + c y ≤ ax2 + bx + c