diketahui p dan q akar akar persamaan [tex] {x}^{2} + (2a + 1)x + 20 = 0[/tex] jika p<0,q<0 dan q-p=1, nilai 2a adalah...

jawab

x² + (2a +1) x + 20 =0
a= 1
b = 2a  + 1
c = 20


p + q = -b/a 
p + q = - (2a +1) .....(i)

p q = c./a 
p q  = 20 --> q = 20/p .....(ii)

q - p = .1 .
20/p - p= 1 ....kalikan p
20 - p² = p
p²  + p - 20 =0
(p + 5)(p -4) =0
p = -5   atau p =  4
p< 0 -->p = - 5

q - p = - 1
q = p - 1
q = - 5 - 1
q = - 6

p + q=  - (2a + 1)
-5 - 6 = - (2a + 1)
11 = 2a + 1
2a = 10
a= 5