Solve the following equation for \( z \) \[ 4 z-1=\sqrt{25 z^{2}-14 z+2} \]

To solve the equation: \[ 4z - 1 = \sqrt{25z^2 - 14z + 2} \] **Step 1: Determine the Domain** Both sides of the equation must be real and non-negative: 1. **Square Root Condition:** \[ 25z^2 - 14z + 2 \geq 0 \] 2. **Left Side Non-Negative:** \[ 4z - 1 \geq 0 \quad \Rightarrow \quad z \geq \frac{1}{4} \] **Step 2: Square Both Sides** To eliminate the square root, square both sides of the equation: \[ (4z - 1)^2 = (\sqrt{25z^2 - 14z + 2})^2 \] \[ 16z^2 - 8z + 1 = 25z^2 - 14z + 2 \] **Step 3: Rearrange the Equation** Bring all terms to one side to form a quadratic equation: \[ 16z^2 - 8z + 1 - 25z^2 + 14z - 2 = 0 \] \[ -9z^2 + 6z - 1 = 0 \] Multiply through by \(-1\) to simplify: \[ 9z^2 - 6z + 1 = 0 \] **Step 4: Solve the Quadratic Equation** Use the quadratic formula: \[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 9 \), \( b = -6 \), and \( c = 1 \): \[ z = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 9 \cdot 1}}{2 \cdot 9} = \frac{6 \pm \sqrt{36 - 36}}{18} = \frac{6}{18} = \frac{1}{3} \] **Step 5: Verify the Solution** Ensure that \( z = \frac{1}{3} \) satisfies the original equation and the domain conditions: 1. \( z = \frac{1}{3} \geq \frac{1}{4} \) ✔️ 2. Substitute \( z = \frac{1}{3} \) into the original equation: \[ 4 \cdot \frac{1}{3} - 1 = \sqrt{25 \cdot \left(\frac{1}{3}\right)^2 - 14 \cdot \frac{1}{3} + 2} \] \[ \frac{4}{3} - 1 = \sqrt{\frac{25}{9} - \frac{14}{3} + 2} \] \[ \frac{1}{3} = \sqrt{\frac{25}{9} - \frac{42}{9} + \frac{18}{9}} = \sqrt{\frac{1}{9}} = \frac{1}{3} \quad ✔️ \] Therefore, the solution is: \[ z = \frac{1}{3} \]