Question

If alpha and beta are the zeroes of 2y2+7y +5, find the value of
(a+ b) + alphabeta.​

Answer 1

2y² + 7y + 5

α + β = -b/a = -7/2

αβ = c/a = 5/2

(α + β)(αβ) = -7/2 × 5/2

= -35/4

hope it helps u..

Answer 2

Hello!

SolutioN :

$\checkmark \:\:\:\sf for \:\: any \:\: quadratic \:\:eq^n\\\sf we\:\:can \:\: say, x^2 -(sum \:\: of \:\:roots)x+product\:\: of \:\:roots=0$

$\sf 2y^2+7y+5=0\\\\\mapsto y^2+\frac{7}{2}y+\frac{5}{2}=0$

$\mapsto\sf y^2-(-\frac{7}{2})y+\frac{5}{2}=0$

$\Boxed \sf hence, \:\: we \:\: we \:\: can \:\:say$

$\sf \alpha+\beta=-\frac{7}{2}\\\sf \&\\\alpha\beta=\frac{5}{2}$

$\rightsquigarrow\boxed{(\alpha+\beta)+\alpha\beta=-1}$

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