Question

alpha beta are the zeros of the quadratic polynomial 2 X square + 5 x + K find the value of k such that alpha + beta whole square minus alpha beta is equal to 24.Its emergency please​

Answer 1

$\huge\boxed{\blue{-71/2}}$

$if \: \alpha \: and \: \beta \: are \: the \: zeros \: of \: {2x }^{2} + 5x + k \\ such \: that \: { (\alpha + \beta) }^{2} - \alpha \beta = 24 \\ \alpha + \beta = \frac{ - b}{a} \\ and \: \alpha \beta = \frac{c}{a} \\ here \: in \: the \: polynomial \: according \: to \: {ax}^{2} + bx + c \\ a = 2 \: \\ b =5 \\ c = k \\ so \: substituting \: the \: values \\ \alpha + \beta = \frac{ - 5 }{2} \: and \: \alpha \beta = \frac{k}{2} \\ substituting \: this \: in \: our \: first \: equation \\ we \: get \: { \frac{ (- 5)}{(2)} }^{2} - \frac{k}{2} = 24 \\ \frac{ 25 - 2k}{4} = 24 \:( because \: we \: take \: lcm)\\ 25 - 2k = 24 \times 4 \\ 25 - 2k = 96 \\ - 2k = 96 - 25 \\ k = \frac{71}{ - 2} \\ thus \: k = - \frac{71}{2}$

$\huge{\purple{Hope\: it \:helps}}$