Eric Nicholas K's answer to 111's Junior College 1 H2 Maths Singapore question.
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This is for part i which is needed for part ii; I will write workings for part ii later
Date Posted:
3 years ago
Alternatively, we can "work backwards" for p and q
x² + px - q = 0 (because y can only equal zero when the numerator is zero but the denominator is non-zero)
From the roots at x = -2 and x = 1,
(x + 2) (x - 1) = 0
x² - x + 2x - 2 = 0
x² + x - 2 = 0
Comparing,
p = 1, q = 2
(note that q = 2 and not q = -2)
x² + px - q = 0 (because y can only equal zero when the numerator is zero but the denominator is non-zero)
From the roots at x = -2 and x = 1,
(x + 2) (x - 1) = 0
x² - x + 2x - 2 = 0
x² + x - 2 = 0
Comparing,
p = 1, q = 2
(note that q = 2 and not q = -2)
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