Question

if alpha and bita are the zeros of 2y square+ 7y+5 . find the value of alpha+beta+alpha×beta

Answer 1

2y^2+7y+5
alpha+bita=-b/a
=-(7)/2=-7/2
alpha ×bita=c/a=5/2
A.T.Q
(Alpha+Bita)+(alpha ×bita)
=(-7/2)+(5/2)
=(-7+5)/2
=-2/2=-1

Answer 2

zeroes of
$2 {y}^{2} + 7y + 5$
will be
$\alpha = \frac{ - b + \sqrt{ {b}^{2} - 4ac }} {2a}$

$\beta = \frac{ - b - \sqrt{ {b}^{2} - 4ac } }{2a}$
so we get
$\alpha = - 1 \\ \beta = - \frac{5}{2}$
so
$( \alpha + \beta ) + ( \alpha \times \beta ) = ( - 1 - \frac{5}{2}) + ( - 1 \times - \frac{5}{2} )$
$= -1$