Jun Heng's answer to lana's Secondary 3 A Maths Singapore question.
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p²x²–(p+2)x+1 = 0
Discriminant > 0
[–(p+2)]²–4(p²)(1) > 0
p²+4p+4–4p² > 0
–3p²+4p+4 > 0
3p²–4p–4 < 0
(3p+2)(p–2) < 0
3p+2 > 0 or p–2 < 0
p > –2/3 or p < 2
–2/3 < p < 2
Discriminant > 0
[–(p+2)]²–4(p²)(1) > 0
p²+4p+4–4p² > 0
–3p²+4p+4 > 0
3p²–4p–4 < 0
(3p+2)(p–2) < 0
3p+2 > 0 or p–2 < 0
p > –2/3 or p < 2
–2/3 < p < 2
Date Posted:
1 year ago
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