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secondary 3 | A Maths
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Date Posted: 5 years ago
Views: 300
(q + 1)x2 + 2x = qx + q - 1
(q + 1)x2 + 2x - qx + 1 - q = 0
(q + 1)x2 + (2 - q)x + 1 - q = 0
“a” = q + 1
“b” = 2 - q
“c” = 1 - q
“b2 - 4ac”
= (2 - q)^2 - 4 (1 + q) (1 - q)
= 4 - 4q + q2 - 4 (1 - q2)
= 4 - 4q + q2 - 4 + 4q2
= 5q2 - 4q
(q + 1)x2 + 2x - qx + 1 - q = 0
(q + 1)x2 + (2 - q)x + 1 - q = 0
“a” = q + 1
“b” = 2 - q
“c” = 1 - q
“b2 - 4ac”
= (2 - q)^2 - 4 (1 + q) (1 - q)
= 4 - 4q + q2 - 4 (1 - q2)
= 4 - 4q + q2 - 4 + 4q2
= 5q2 - 4q
Of course 5q2 - 4q = q (5q - 4)
When q is negative, so does 5q - 4
The product of two negative numbers is positive
So b2 - 4ac is positive when q is negative, hence the curve has real and distinct roots for all negative values of q
When q is negative, so does 5q - 4
The product of two negative numbers is positive
So b2 - 4ac is positive when q is negative, hence the curve has real and distinct roots for all negative values of q
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Dk if u still need it but here you go
Date Posted:
5 years ago
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