mohon bantuannya kakak2 sekalian terima kasih

jawab

PK1 
x² - 5x - 7 = 0 akr akarnya  αdan β
a+b = 5
ab = - 7
PK2 akarnya p=  2/(α+2) dan q  = 2/(β+2)

p + q = 2/(a+2) + 2/(b+2)
p + q = 2(b+2)+2(a+2) / (a +2)(b+2)
p+ q = (2b +4 +2a+ 4)/(a+2)(b+2)
p + q = 2(a + b) + 8)/(ab +2ab + 4
p+ q =  (2(5) + 8))/ (-7 + 2(5) + 4)
p + q = 18/7

p q =  2/(a+2) . 2/(b+2)
p q = 4/(ab + 2ab + 4) = 4/(-7 +2(4) + 4))= 4/7

PK 2
x²  - (p+q)x + p. q =0
x² - 18./7 x + 4/7 =0
7x² -18x + 4=  7

b)
p=  2 + 1/a
q = 2 + 1/b
p+q = 4 + 1/a +1/b = 4 + (a+b)/(ab) = 4 + 5/-7 = 23/7
pq = (2 + 1/a)(2+1/b) = 4 + 2(1/a+1/b) + 1/(ab)
p q = 4 + 2(a+b)/(ab) + 1/(ab)
p q = 4 + 2(5)/-7 + 1/-7
p q = 4 - 11/7 = 17/7
PK2 
x² - (p+q)x + (p q) = 0
x²- (23/7)x + (17/7) =0
7x² -23x + 17 =0

Kategori soal : matematika
Kelas : 10 SMA
Materi : persamaan kuadrat
Kata kunci : akar-akar persamaan

Pembahasan : perhitungan terlampir