jawab no 32 dengan benar

smoga bermanfaat...............

Mapel : Matematika
Materi : Persamaan Kuadrat
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x² - 2x - 4 = 0

a = 1
b = -2
c = -4

X12 = [-b ± √b² - 4ac] / 2a

X12 = [-(-2) ± √(-2)² - 4(1)(-4)] / 2

X12 = [2 ± √4 + 16] /2

X12 = [2 ± √20] / 2

X12 = [2 ± 2√5] / 2

X12 = 1 ± √5

→ x1 = 1 + √5
→ x2 = 1 - √5

Sehingga, akar-akar persamaan baru :
P = x1 - 2
P = 1 + √5 - 2
P = √5 - 1

Q = x2 - 2
Q = 1 - √5 - 2
Q = -1 - √5

Nilai persamaan kuadrat barunya adalah

x² - (P + Q) x + P•Q = 0

x² - [√5 - 1 + (-1 - √5)] x + [(√5 -1)(-1 -√5)] = 0

x² - (-2)x + (-√5 - 5 + 1 + √5) = 0

x² + 2x + (-4) = 0

x² + 2x - 4 = 0

Semoga membantu, maaf jika ada salah..