one root of the equation \( 2 x^{2}+5 x+k=0 \) is reciprocal to each other, then \( k \) is

Given the quadratic equation: \[ 2x^2 + 5x + k = 0 \] Let's assume the roots of this equation are \( r \) and \( \frac{1}{r} \) since one root is the reciprocal of the other. **Using Vieta's Formulas:** 1. **Sum of the roots:** \[ r + \frac{1}{r} = -\frac{b}{a} = -\frac{5}{2} \] 2. **Product of the roots:** \[ r \cdot \frac{1}{r} = \frac{c}{a} = \frac{k}{2} \] \[ 1 = \frac{k}{2} \implies k = 2 \] **Verification:** To ensure that the roots are real, the discriminant must be non-negative: \[ \Delta = b^2 - 4ac = 5^2 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9 \geq 0 \] Since the discriminant is positive, the equation has two distinct real roots, and \( k = 2 \) is valid. **Answer:** \( \boxed{2} \)