Question

if alpha and beeta are zeores of polynomial 2x^2+5x+k find k such that (alpha +beeta)^2-alpha×beeta=24


handwritten answers plz
..very urgent ​

Answer 1

PLEASE MARK IT AS BRAINLIEST AND FOLLOW ME

Answer:

-23 is the answer

Step-by-step explanation:

$2 {x}^{2} + 5x + k$

Zeroes are -

$\alpha \: and \: \beta$

$( { \alpha + \beta })^{2} - \alpha \beta = 24$

$\alpha + \beta = - \frac{b}{a} = - \frac{5}{2}$

$\alpha \beta = \frac{c}{a} = \frac{k}{2}$

$put \: the \: \: values \: which \: we \: get \: earlier$

$( - { \frac{5}{2} })^{2} - \frac{k}{2} = 24$

$\frac{25}{2} - \frac{k}{2} = 24$

$\frac{25 - k}{2} = 24$

$25 - k = 48$

$k = - 23$