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secondary 3 | A Maths
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Date Posted: 2 years ago
Views: 198
x² + (p+1)∙x + (2p - 5) = 0
a = 1
b = p + 1
c = 2p - 5
b² - 4ac
= (p + 1)² - 4∙(1p - 5)
= p² - 2p + 25
= (p - 1)² + 24 > 0
a = 1
b = p + 1
c = 2p - 5
b² - 4ac
= (p + 1)² - 4∙(1p - 5)
= p² - 2p + 25
= (p - 1)² + 24 > 0
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done
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We need to show that discriminant of roots of the quadratic equation is always zero or greater than zero to conclude that the quadratic equation will have real roots.
The idea of the discriminant of roots of a quadratic equation actually has its origins from one particular expression in the quadratic formula.
The idea of the discriminant of roots of a quadratic equation actually has its origins from one particular expression in the quadratic formula.
Date Posted:
2 years ago
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