Math Problem Statement

Solve each non-linear inequality. Write the final answer in interval notation.

x2 – 2x > 35 2x2 + 7x – 4 ≥ 0 2x2 – 5x < – 3

Solution

Let's solve each inequality step-by-step and write the solutions in interval notation.

1. Solve:

x22x>35x^2 - 2x > 35

Step 1: Move all terms to one side of the inequality.

x22x35>0x^2 - 2x - 35 > 0

Step 2: Factor the quadratic expression.

We need to find two numbers that multiply to 35-35 and add to 2-2. The numbers are 7-7 and 55. (x7)(x+5)>0(x - 7)(x + 5) > 0

Step 3: Determine the critical points.

The critical points are x=7x = 7 and x=5x = -5.

Step 4: Analyze the sign of the expression around the critical points.

We now check the sign of (x7)(x+5)(x - 7)(x + 5) in the intervals (,5)(-\infty, -5), (5,7)(-5, 7), and (7,)(7, \infty).

  • For x(,5)x \in (-\infty, -5), pick x=6x = -6: (67)(6+5)=(13)(1)>0(-6 - 7)(-6 + 5) = (-13)(-1) > 0
  • For x(5,7)x \in (-5, 7), pick x=0x = 0: (07)(0+5)=(7)(5)<0(0 - 7)(0 + 5) = (-7)(5) < 0
  • For x(7,)x \in (7, \infty), pick x=8x = 8: (87)(8+5)=(1)(13)>0(8 - 7)(8 + 5) = (1)(13) > 0

Step 5: Write the solution.

The expression is positive when x(,5)(7,)x \in (-\infty, -5) \cup (7, \infty).

Final answer: (,5)(7,)(-\infty, -5) \cup (7, \infty)


2. Solve:

2x2+7x402x^2 + 7x - 4 \geq 0

Step 1: Factor the quadratic expression.

We factor 2x2+7x4=(2x1)(x+4)2x^2 + 7x - 4 = (2x - 1)(x + 4).

Step 2: Determine the critical points.

Set each factor equal to zero: 2x1=0orx+4=02x - 1 = 0 \quad \text{or} \quad x + 4 = 0 Thus, the critical points are x=12x = \frac{1}{2} and x=4x = -4.

Step 3: Analyze the sign of the expression around the critical points.

We check the sign of (2x1)(x+4)(2x - 1)(x + 4) in the intervals (,4)(-\infty, -4), (4,12)(-4, \frac{1}{2}), and (12,)(\frac{1}{2}, \infty).

  • For x(,4)x \in (-\infty, -4), pick x=5x = -5: (2(5)1)(5+4)=(11)(1)>0(2(-5) - 1)(-5 + 4) = (-11)(-1) > 0
  • For x(4,12)x \in (-4, \frac{1}{2}), pick x=0x = 0: (2(0)1)(0+4)=(1)(4)<0(2(0) - 1)(0 + 4) = (-1)(4) < 0
  • For x(12,)x \in (\frac{1}{2}, \infty), pick x=1x = 1: (2(1)1)(1+4)=(1)(5)>0(2(1) - 1)(1 + 4) = (1)(5) > 0

Step 4: Write the solution, including the points where the expression equals zero.

The expression is non-negative when x(,4][12,)x \in (-\infty, -4] \cup [\frac{1}{2}, \infty).

Final answer: (,4][12,)(-\infty, -4] \cup [\frac{1}{2}, \infty)


3. Solve:

2x25x<32x^2 - 5x < -3

Step 1: Move all terms to one side of the inequality.

2x25x+3<02x^2 - 5x + 3 < 0

Step 2: Factor the quadratic expression.

We need to factor 2x25x+3=(2x3)(x1)2x^2 - 5x + 3 = (2x - 3)(x - 1).

Step 3: Determine the critical points.

Set each factor equal to zero: 2x3=0orx1=02x - 3 = 0 \quad \text{or} \quad x - 1 = 0 Thus, the critical points are x=32x = \frac{3}{2} and x=1x = 1.

Step 4: Analyze the sign of the expression around the critical points.

We check the sign of (2x3)(x1)(2x - 3)(x - 1) in the intervals (,1)(-\infty, 1), (1,32)(1, \frac{3}{2}), and (32,)(\frac{3}{2}, \infty).

  • For x(,1)x \in (-\infty, 1), pick x=0x = 0: (2(0)3)(01)=(3)(1)>0(2(0) - 3)(0 - 1) = (-3)(-1) > 0
  • For x(1,32)x \in (1, \frac{3}{2}), pick x=1.25x = 1.25: ((2(1.25) - 3)(1.25 - 1

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Inequalities
Factoring
Critical Points
Interval Notation

Formulas

Quadratic equation: ax^2 + bx + c = 0
Factoring of quadratic expressions

Theorems

Sign Analysis for Quadratic Inequalities
Zero Product Property

Suitable Grade Level

Grades 9-12