Tentukan akar² pers. kuadrat berikut dengan mengungkapkan kuadrat sempurna:
1. X²+6x-16=0
2. X²-4x-45=0
Jawab:
(1.) Hp = {2, -8}
(2.) Hp = {9, -5}
Penjelasan :
(1.) x² + 6x – 16 = 0
x² + 6x = 16
x² + 6x + (⁶/₂)² = 16 + (⁶/₂)²
x² + 6x + 3² = 16 + 3²
x² + 2x(3) + 3² = 16 + 9
∵ a²+2ab+b² = (a+b)² ∴
(x+3)² = 25
x+3 = ±√25
x + 3 = ±5
x = ±5 – 3
Hp = {5-3, -5-3}
Hp = {2, -(5+3)}
Hp = {2, -8}
———————–
(2.) x² – 4x – 45 = 0
x² – 4x = 45
x² – 4x + (⁻⁴/₂)² = 45 + (⁻⁴/₂)²
x² – 4x + (-2)² = 45 + (-2)²
x² – 2x(2) + 2² = 45 + 4
∵ a²-2ab+b² = (a-b)² ∴
(x – 2)² = 49
x-2 = ±√49
x – 2 = ±7
x = ±7 + 2
Hp = {7+2, -7+2}
Hp = {9, -(7-2)}
Hp = {9, -5}
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