# 34 & # 40 algebra help?
回答 (1)
2015-09-09 1:29 pm
34).
√(2n + 29) + 3 = n
(2n + 29) = (n - 3)^2
n^2 - 8n - 20 = 0
(n - 10)(n + 2) = 0
Solutions:
n = -2
n = 10
40).
√(d + 4) - √(6 + 2d) = -1
√(6 + 2d) - √(d + 4) = 1
3d + 10 = 1 + 2√(6 + 2d)√(d + 4)
Solution:
d = 5
√(2n + 29) + 3 = n
(2n + 29) = (n - 3)^2
n^2 - 8n - 20 = 0
(n - 10)(n + 2) = 0
Solutions:
n = -2
n = 10
40).
√(d + 4) - √(6 + 2d) = -1
√(6 + 2d) - √(d + 4) = 1
3d + 10 = 1 + 2√(6 + 2d)√(d + 4)
Solution:
d = 5
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https://hk.answers.yahoo.com/question/index?qid=20150909030810AAGnUEJ