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: \[ (2, 3) \]