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secondary 3 | E Maths
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Date Posted: 1 year ago
Views: 270
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(x-(-3))(x-5) = 0
(x+3)(x-5) = 0
x^2 + 3x - 5x - 15 = 0
x^2 - 2x - 15 = 0
By comparing coefficients,
b = -2, c = -15
(x+3)(x-5) = 0
x^2 + 3x - 5x - 15 = 0
x^2 - 2x - 15 = 0
By comparing coefficients,
b = -2, c = -15
I need the working for part b
i added it alr
b) (x-(-p))(x-q)=0
(x+p)(x-q) = 0
x^2 + px -qx -pq= 0
x^2 + (p-q)x - pq = 0
b = p-q
c = -pq
(x+p)(x-q) = 0
x^2 + px -qx -pq= 0
x^2 + (p-q)x - pq = 0
b = p-q
c = -pq
Cannot undertsand how to get to (x-(-p))(x-q)=0
done
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clear
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Part b only
Date Posted:
1 year ago
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