If one zero is the reciprocal …

The product of the zeros of a quadratic equation is equal to the constant term divided by the coefficient of the squared term. So, for the quadratic equation 3kx² - 7x + 11, the product of the zeros would be: x * (1/x) = 1 Now, we can use the sum of the zeros of a quadratic equation, which is equal to the negative coefficient of the linear term divided by the coefficient of the squared term. So, for our equation: x + (1/x) = -(-7) / 3k x + (1/x) = 7 / 3k Now, we have two equations: x * (1/x) = 1 x + (1/x) = 7 / 3k Let's simplify the first equation: x * (1/x) = 1 1 = 1 Now, for the second equation, let's solve for x + (1/x): x + (1/x) = 7 / 3k Since we know that x * (1/x) = 1, we can multiply both sides of the second equation by x: x * (x + (1/x)) = x * (7 / 3k) This simplifies to: x² + 1 = 7x / 3k Now, let's find the common denominator on the right side: x² + 1 = (7x * x) / (3k * x) x² + 1 = (7x²) / (3kx) Now, let's bring all terms to one side of the equation: x² + 1 - (7x²) / (3kx) = 0 To solve this quadratic equation for x, we need to find the common denominator: (3kx * x² + 3kx - 7x²) / (3kx) = 0 Now, we need to find the LCD (Least Common Denominator) of the terms in the numerator: LCD = 3kx Now, let's multiply both sides of the equation by the LCD to get rid of the fraction: 3kx * (3kx * x² + 3kx - 7x²) / (3kx) = 0 * (3kx) This simplifies to: 3kx * (3kx * x² + 3kx - 7x²) = 0 Expanding the expression: 9k²x³ + 3k²x² - 21kx³ = 0 Combine the x³ terms: (9k² - 21k) * x³ + 3k²x² = 0 For this equation to be true for all x, the coefficients of x³ and x² must be zero: 9k² - 21k = 0 (Coefficient of x³) 3k² = 0 (Coefficient of x²) Solving the second equation: 3k² = 0 k² = 0 k = 0 Now, for the first equation: 9k² - 21k = 0 9(0)² - 21(0) = 0 0 - 0 = 0 Since both equations are true when k = 0, the value of k that satisfies the conditions is k = 0.