2x^2-7x+3<0 X^2-2x>0

To solve the inequalities \(2x^2 - 7x + 3 < 0\) and \(x^2 - 2x > 0\), we'll handle each one separately.

Step 1: Solve \(2x^2 - 7x + 3 < 0\)

First, find the roots of the equation \(2x^2 - 7x + 3 = 0\) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \(a = 2\), \(b = -7\), and \(c = 3\).

Calculating the discriminant:

\[ b^2 - 4ac = (-7)^2 - 4(2)(3) = 49 - 24 = 25 \]

Now substituting in the quadratic formula:

\[ x = \frac{7 \pm \sqrt{25}}{2(2)} = \frac{7 \pm 5}{4} \]

This gives us two solutions:

1. \(x = \frac{12}{4} = 3\)
2. \(x = \frac{2}{4} = \frac{1}{2}\)

Now we have roots \(x = \frac{1}{2}\) and \(x = 3\).

Testing intervals for \(2x^2 - 7x + 3 < 0\)

The intervals to test are:

• \( (-\infty, \frac{1}{2}) \)
• \( \left(\frac{1}{2}, 3\right) \)
• \( (3, \infty) \)

1. Test \(x = 0\) in \((- \infty, \frac{1}{2})\): \[ 2(0)^2 - 7(0) + 3 = 3 > 0 \quad \text{(not in solution)} \]

2. Test \(x = 1\) in \(\left(\frac{1}{2}, 3\right)\): \[ 2(1)^2 - 7(1) + 3 = 2 - 7 + 3 = -2 < 0 \quad \text{(in solution)} \]

3. Test \(x = 4\) in \((3, \infty)\): \[ 2(4)^2 - 7(4) + 3 = 32 - 28 + 3 = 7 > 0 \quad \text{(not in solution)} \]

Thus, the solution for \(2x^2 - 7x + 3 < 0\) is: \[ \left(\frac{1}{2}, 3\right) \]

Step 2: Solve \(x^2 - 2x > 0\)

Factoring the expression gives us: \[ x(x - 2) > 0 \]

Finding the roots:

• \(x = 0\)
• \(x = 2\)

Testing intervals for \(x^2 - 2x > 0\)

The intervals to test are:

• \( (-\infty, 0) \)
• \( (0, 2) \)
• \( (2, \infty) \)

1. Test \(x = -1\) in \((- \infty, 0)\): \[ (-1)(-1 - 2) = (-1)(-3) = 3 > 0 \quad \text{(in solution)} \]

2. Test \(x = 1\) in \((0, 2)\): \[ (1)(1 - 2) = (1)(-1) = -1 < 0 \quad \text{(not in solution)} \]

3. Test \(x = 3\) in \((2, \infty)\): \[ (3)(3 - 2) = (3)(1) = 3 > 0 \quad \text{(in solution)} \]

Thus, the solution for \(x^2 - 2x > 0\) is: \[ (-\infty, 0) \cup (2, \infty) \]

Step 3: Combine the solutions

We now need to find the intersection of the solutions:

1. From the first inequality: \(\left(\frac{1}{2}, 3\right)\)
2. From the second inequality: \( (-\infty, 0) \cup (2, \infty) \)

The intersection is found within the interval \(\left(\frac{1}{2}, 3\right)\):

• The only overlapping portion is \( (2, 3) \).

Final Solution

The final solution to the system of inequalities is: \[ (2, 3) \]