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

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

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

### 1. Solve: 
\[
x^2 - 2x > 35
\]
#### Step 1: Move all terms to one side of the inequality.
\[
x^2 - 2x - 35 > 0
\]

#### Step 2: Factor the quadratic expression.
We need to find two numbers that multiply to \(-35\) and add to \(-2\). The numbers are \(-7\) and \(5\).
\[
(x - 7)(x + 5) > 0
\]

#### Step 3: Determine the critical points.
The critical points are \(x = 7\) and \(x = -5\).

#### Step 4: Analyze the sign of the expression around the critical points.
We now check the sign of \((x - 7)(x + 5)\) in the intervals \((-\infty, -5)\), \((-5, 7)\), and \((7, \infty)\).

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

#### Step 5: Write the solution.
The expression is positive when \(x \in (-\infty, -5) \cup (7, \infty)\).

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

---

### 2. Solve: 
\[
2x^2 + 7x - 4 \geq 0
\]

#### Step 1: Factor the quadratic expression.
We factor \(2x^2 + 7x - 4 = (2x - 1)(x + 4)\).

#### Step 2: Determine the critical points.
Set each factor equal to zero:
\[
2x - 1 = 0 \quad \text{or} \quad x + 4 = 0
\]
Thus, the critical points are \(x = \frac{1}{2}\) and \(x = -4\).

#### Step 3: Analyze the sign of the expression around the critical points.
We check the sign of \((2x - 1)(x + 4)\) in the intervals \((-\infty, -4)\), \((-4, \frac{1}{2})\), and \((\frac{1}{2}, \infty)\).

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

#### Step 4: Write the solution, including the points where the expression equals zero.
The expression is non-negative when \(x \in (-\infty, -4] \cup [\frac{1}{2}, \infty)\).

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

---

### 3. Solve:
\[
2x^2 - 5x < -3
\]

#### Step 1: Move all terms to one side of the inequality.
\[
2x^2 - 5x + 3 < 0
\]

#### Step 2: Factor the quadratic expression.
We need to factor \(2x^2 - 5x + 3 = (2x - 3)(x - 1)\).

#### Step 3: Determine the critical points.
Set each factor equal to zero:
\[
2x - 3 = 0 \quad \text{or} \quad x - 1 = 0
\]
Thus, the critical points are \(x = \frac{3}{2}\) and \(x = 1\).

#### Step 4: Analyze the sign of the expression around the critical points.
We check the sign of \((2x - 3)(x - 1)\) in the intervals \((-\infty, 1)\), \((1, \frac{3}{2})\), and \((\frac{3}{2}, \infty)\).

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

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

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

Explore how to solve non-linear quadratic inequalities such as x^2 – 2x > 35, 2x^2 + 7x – 4 ≥ 0, and 2x^2 – 5x < – 3. This solution covers key algebraic techniques including factoring, critical points, and interval notation, suitable for high school students in Grades 9-12.

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