If (a+b)^1/2 = (a-b)^-1/2(negative), which of the following must be true?
A. b = 0
B. a+b = 1
C. a-b = 1
D. a^2 + b^2 = 1
E. a^2 - b^2 = 1
2007-09-30 21:21:54 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
E, assuming a<>b
2007-09-30 21:27:26 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The correct answer is E. Because...
(a^2 - b^2) = 1
(difference of 2 squares)
(a+b)(a-b) = 1
a+b = 1/(a-b)
so,
(a+b)^1/2 = (1/(a-b))^1/2
=(a-b)^ -1/2
2007-10-01 04:30:45 · answer #2 · answered by mevelyn2551 3 · 0⤊ 0⤋
E
(a+b)^1/2 = (a-b)^-1/2
((a+b)(a-b))^1/2 = 1
so a^2 +ab -ab -b^2 = 1
a^2 - b^2 = 1
2007-10-01 04:29:39 · answer #3 · answered by Steve 2 · 0⤊ 0⤋
(a+b)^(1/2) = (a-b)^(-1/2)
square both side:
[(a+b)^(1/2)]^2 = [(a-b)^(-1/2)]^2
a + b = 1/ (a-b)
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cross multiplication:
(a+b) (a-b) = 1
a^2 - ab + ab - b^2 = 1
a^2 - b^2 = 1
so the answer is e
2007-10-01 04:30:16 · answer #4 · answered by UJ 3 · 0⤊ 0⤋
(a-b)^-1/2 =1/(a-b)^1/2
(a+b)^1/2= 1/(a-b)^1/2
(a+b)^1/2*(a-b)^1/2=1 I square this equation
(a+b)(a-b) =1
a^2-b^2 =1
Answer E
2007-10-01 04:29:05 · answer #5 · answered by maussy 7 · 0⤊ 0⤋
(a+b)^1/2 = (a-b)^-1/2
(a+b) = ((a-b)^-1/2)^2
(a+b) = (a-b)^-1
(a+b) = 1/(a-b)
(a+b)(a-b) = 1
a^2 +ab - ab - b^2 = 1
a^2 - b^2 = 1 <----- E.
2007-10-01 04:35:59 · answer #6 · answered by Holden 5 · 0⤊ 0⤋
Square bot sides, we have
a+b = (a-b)^-1
multiply both by a-b
(a+b)(a-b) = 1
so the answer is E
.
2007-10-01 04:27:29 · answer #7 · answered by tsr21 6 · 0⤊ 0⤋