persamaan kuadrat x^2+6x+10=0 mempunyai akar akar p dan q. persamaan kuadrat yang akar akarnya 2p^2 dan 2q^2 adalah

p + q = -6
pq = 10

2p^2*2q^2 = 4p^2q^2
= 4(10)(10)
= 400 = c/a

2p^2 + 2q^2 = 2(p^2 + q^2)
= 2((p+q)^2 - 2pq)
= 2((-6)^2 - 20)
= 2(36-20)
=2(16)
= 32 = -b/a
b = -32

x^2 - 32x + 400 = 0

x^2 + 6x + 10 = 0
p + q = -b/a = -6
p.q = c/a = 10
Misal x1 = 2p^2 dan x2 = 2q^2

x1 + x2 = 2p^2 + 2q^2 = 2(p^2 + q^2) = 2((p + q)^2 - 2pq) = 2((-6)^2 - 2(10))
= 2(36 - 20) = 2(16) = 32

x1.x2 = (2p^2)(2q^2) = 4(pq)^2 = 4(10)^2 = 400

Persamaan kuadrat baru
x^2 - (x1 + x2)x + x1x2 = 0
x^2 - 32x + 400 = 0