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secondary 3 | E Maths
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Date Posted: 4 years ago
Views: 258
Part a is one of the cases where we do not do a two-by-two grouping, but rather a three-by-one grouping.
p8 - 4p4 + 4 - 16q2r2
= [p8 - 4p4 + 4] - 16q2r2
= (p4 - 2) (p4 - 2) - (4qr)2
= (p4 - 2)2 - (4qr)2
[recall A2 - B2 = (a plus b) (a minus b)]
= (p4 - 2 PLUS 4qr) (p4 - 2 MINUS 4qr)
= (p4 - 2 + 4qr) (p4 - 2 - 4qr)
= [p8 - 4p4 + 4] - 16q2r2
= (p4 - 2) (p4 - 2) - (4qr)2
= (p4 - 2)2 - (4qr)2
[recall A2 - B2 = (a plus b) (a minus b)]
= (p4 - 2 PLUS 4qr) (p4 - 2 MINUS 4qr)
= (p4 - 2 + 4qr) (p4 - 2 - 4qr)
See 1 Answer
done
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Part B. Hope this helps
Date Posted:
4 years ago
mistake in line 6, left side …
should be c^2 instead of c
also, at end, should consider both plus & minus roots
final answer
b=(plus/minus) sqrt[ (a-2ac^2) / (5c^2+1) ]
should be c^2 instead of c
also, at end, should consider both plus & minus roots
final answer
b=(plus/minus) sqrt[ (a-2ac^2) / (5c^2+1) ]
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