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secondary 3 | A Maths
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Date Posted: 2 years ago
Views: 436
a = 1
b = 3k+5
c = -3
D² = b² - 4ac
= (3k+5)² - 4×1×(-3)
= (3k+5)² + 12
= X + Y > 0
X → positive w k is any real constant
Y → positive
∴ the equation has the real and distinct roots
b = 3k+5
c = -3
D² = b² - 4ac
= (3k+5)² - 4×1×(-3)
= (3k+5)² + 12
= X + Y > 0
X → positive w k is any real constant
Y → positive
∴ the equation has the real and distinct roots
See 1 Answer
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Hope this helps!
Date Posted:
2 years ago
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