(5. If one zero of the quadratic polynomial \( f(x)=4 x^{2}-8 k x-9 \) is negative of the other, find the value of \( k \). 6. If the sum of the zeros of the quadratic polynomial \( f(t)=k t^{2}+2 t+3 k \) is equal to their

Let's solve each part of the problem step by step. ### Part 5: Finding the value of \( k \) We are given the quadratic polynomial: \[ f(x) = 4x^2 - 8kx - 9 \] We know that if one zero is the negative of the other, we can denote the zeros as \( r \) and \( -r \). The sum of the zeros \( r + (-r) = 0 \) and the product of the zeros \( r \cdot (-r) = -r^2 \). Using Vieta's formulas, we know: 1. The sum of the zeros \( r + (-r) = \frac{-b}{a} = \frac{8k}{4} = 2k \). 2. The product of the zeros \( r \cdot (-r) = \frac{c}{a} = \frac{-9}{4} \). Since the sum of the zeros is zero, we can set up the equation: \[ 2k = 0 \implies k = 0 \] Now, let's check the product of the zeros: \[ -r^2 = \frac{-9}{4} \implies r^2 = \frac{9}{4} \implies r = \pm \frac{3}{2} \] Thus, the value of \( k \) is: \[ \boxed{0} \] ### Part 6: Finding \( k \) when the sum of the zeros equals their product We are given the quadratic polynomial: \[ f(t) = kt^2 + 2t + 3k \] Let the zeros be \( \alpha \) and \( \beta \). According to Vieta's formulas: 1. The sum of the zeros \( \alpha + \beta = \frac{-b}{a} = \frac{-2}{k} \). 2. The product of the zeros \( \alpha \beta = \frac{c}{a} = \frac{3k}{k} = 3 \). We need to set the sum equal to the product: \[ \frac{-2}{k} = 3 \] Now, let's solve for \( k \): \[ -2 = 3k \implies k = \frac{-2}{3} \] Thus, the value of \( k \) is: \[ \boxed{-\frac{2}{3}} \]