Jawab:
Jarak sumbu simetri
(Jarak antar Xp dari P(x) dan Q(x))
beda 2a satuan.. ✅

Penjelasan dengan langkah-langkah:
Dua fungsi kuadrat P(x) dan Q(x)
P(x) = 0   x_{1p, 2p} = a±√b
x = a±√b
x-a = ±√b
(x-a)² = b
x²-2ax+a² = b
-->  x²-2ax+a²-b = 0
d = 1, e = -2a, f = a²-b
Cari sumbu simetri P(x)
x verteks = -e/2d = -(-2a)/(2(1)) = 2a/2 = a
---------------------------------------
Q(x) = 0   x_{1q, 2q} = -a±√b
x = -a±√b
x+a = ±√b
(x+a)² = b
-->  x²+2ax+a²-b = 0
d = 1, e = 2a, f = a²-b
Cari sumbu simetri Q(x)
x verteks = -e/2d = -(2a)/(2(1)) = -2a/2 = -a
------------------------------------------------------------
x verteks P(x)  xp_{P(x)} = a
x verteks Q(x) xp_{Q(x)} = -a
∵ xp_{P(x)}  >  xp_{Q(x)} ∴
Jarak = xp_{P(x)} - (xp_{Q(x)})
= a-(-a) = a+a = 2a satuan ✅

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