\( \left\{\begin{array}{l}if\alpha and\beta aretherootsof \\4z^{2}-17z+k=0 \\satisfiestherelation \\2\alpha +3\beta =35 \\thenfind \\k\end{array}\right. \)

To find the value of \( k \), we can use Vieta's formulas which state that for a quadratic equation of the form \( az^2 + bz + c = 0 \), the sum of the roots \( \alpha + \beta = -\frac{b}{a} \) and the product of the roots \( \alpha \beta = \frac{c}{a} \). For the given equation \( 4z^2 - 17z + k = 0 \): 1. The sum of the roots \( \alpha + \beta = \frac{17}{4} \). 2. The product of the roots \( \alpha \beta = \frac{k}{4} \). From the relation \( 2\alpha + 3\beta = 35 \), we can express \( \beta \) in terms of \( \alpha \): \[ \beta = \frac{35 - 2\alpha}{3} \] Substituting this expression for \( \beta \) into the sum of the roots equation: \[ \alpha + \frac{35 - 2\alpha}{3} = \frac{17}{4} \] Multiplying through by 3 to eliminate the fraction yields: \[ 3\alpha + 35 - 2\alpha = \frac{51}{4} \] This simplifies to: \[ \alpha + 35 = \frac{51}{4} \] \[ \alpha = \frac{51}{4} - 35 \] \[ \alpha = \frac{51}{4} - \frac{140}{4} = \frac{-89}{4} \] Now using \( \alpha + \beta = \frac{17}{4} \) to find \( \beta \): \[ \beta = \frac{17}{4} - \alpha = \frac{17}{4} + \frac{89}{4} = \frac{106}{4} = \frac{53}{2} \] Now we can find \( k \) using the product of the roots: \[ \alpha \beta = \left(\frac{-89}{4}\right) \left(\frac{53}{2}\right) = \frac{-89 \times 53}{8} \] \[ \alpha \beta = \frac{-4717}{8} \] Now, using \( \alpha \beta = \frac{k}{4} \): \[ \frac{-4717}{8} = \frac{k}{4} \] Cross-multiplying gives: \[ k = \frac{-4717}{8} \times 4 = \frac{-4717}{2} \] Thus, the value of \( k \) is \[ \boxed{\frac{-4717}{2}}. \]