Math Problem Statement

Solve the following inequalities: (a) x^2 - 7x - 8 ≥ 0 (b) x^2 - 4x ≥ 12 (c) x(4x + 1) ≤ 3x

Solution

The image contains the following inequalities that need to be solved:

(a) x27x80x^2 - 7x - 8 \geq 0

(b) x24x12x^2 - 4x \geq 12

(c) x(4x+1)3xx(4x + 1) \leq 3x

Let’s solve these one by one.

(a) x27x80x^2 - 7x - 8 \geq 0

  1. Factor the quadratic expression: x27x8=(x8)(x+1)x^2 - 7x - 8 = (x - 8)(x + 1) So, the inequality becomes: (x8)(x+1)0(x - 8)(x + 1) \geq 0

  2. Identify the critical points: x=8x = 8 and x=1x = -1.

  3. Test intervals:

    • For x<1x < -1, both factors are negative, so the product is positive.
    • For 1<x<8-1 < x < 8, one factor is positive, and one is negative, so the product is negative.
    • For x>8x > 8, both factors are positive, so the product is positive.
  4. The solution to the inequality is x(,1][8,)x \in (-\infty, -1] \cup [8, \infty).


(b) x24x12x^2 - 4x \geq 12

  1. Rearrange the inequality: x24x120x^2 - 4x - 12 \geq 0

  2. Factor or use the quadratic formula: x=(4)±(4)24(1)(12)2(1)=4±16+482=4±642=4±82x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-12)}}{2(1)} = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm \sqrt{64}}{2} = \frac{4 \pm 8}{2} This gives x=6x = 6 and x=2x = -2.

  3. Test intervals:

    • For x<2x < -2, the quadratic expression is positive.
    • For 2<x<6-2 < x < 6, the quadratic expression is negative.
    • For x>6x > 6, the quadratic expression is positive.
  4. The solution is x(,2][6,)x \in (-\infty, -2] \cup [6, \infty).


(c) x(4x+1)3xx(4x + 1) \leq 3x

  1. Move all terms to one side: x(4x+1)3x0x(4x + 1) - 3x \leq 0 Simplify: 4x2+x3x04x22x04x^2 + x - 3x \leq 0 \quad \Rightarrow \quad 4x^2 - 2x \leq 0

  2. Factor the quadratic: 2x(2x1)02x(2x - 1) \leq 0

  3. Identify critical points: x=0x = 0 and x=12x = \frac{1}{2}.

  4. Test intervals:

    • For x<0x < 0, the expression is positive.
    • For 0<x<120 < x < \frac{1}{2}, the expression is negative.
    • For x>12x > \frac{1}{2}, the expression is positive.
  5. The solution is x[0,12]x \in [0, \frac{1}{2}].


Final Solutions:

  • (a) x(,1][8,)x \in (-\infty, -1] \cup [8, \infty)
  • (b) x(,2][6,)x \in (-\infty, -2] \cup [6, \infty)
  • (c) x[0,12]x \in [0, \frac{1}{2}]

Would you like further details on any of these steps or explanations?

Here are 5 related questions for further understanding:

  1. How do you solve inequalities using factorization?
  2. What is the importance of testing intervals in inequalities?
  3. How does the quadratic formula help in solving inequalities?
  4. What happens if you multiply or divide inequalities by a negative number?
  5. Can inequalities have multiple solution intervals?

Tip: Always test values from different intervals to check the sign of the expression when solving inequalities.

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Math Problem Analysis

Mathematical Concepts

Algebra
Inequalities
Quadratic Equations

Formulas

Quadratic Formula: x = [-b ± √(b² - 4ac)] / 2a

Theorems

Zero Product Property
Interval Testing Method

Suitable Grade Level

Grades 9-10