Question

jika persamaan kuadrat x^2 + px + q = 0 mempunyai dua akar real , maka ....
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Answer 1

x² + px + q = 0

a = 1

b = p

c = q

D = b² - 4ac

D = p² - 4(1)(q)

D = p² - 4(q)

D = p² - 4q

Syarat Akar Real = D > 0

p² - 4q > 0

p² > 0 + 4q

p² > 4q [ A. ]

:)