if the sum of the reciprocals of two consecutive odd integers is 12/35
Answers
Answered by
15
let first no . be x
then
1/x+1/(x+2)=12/35
or (2x+2)/x²+2x=12/35
or 70x+70=12x²+24x
or 12x²-46x-70=0
or 6x²-23x-35=0
or 6x²-30x+7x-35=0
or (x-5)(5x+7)=0
:. x=5 or x=-7/5
so answer is 5.
then
1/x+1/(x+2)=12/35
or (2x+2)/x²+2x=12/35
or 70x+70=12x²+24x
or 12x²-46x-70=0
or 6x²-23x-35=0
or 6x²-30x+7x-35=0
or (x-5)(5x+7)=0
:. x=5 or x=-7/5
so answer is 5.
Answered by
8
Thank you for asking this question. The options for this question are missing, here are the missing options:
A. 3
B. 5
C. 7
D. 9
E. 11
Answer:
We will let n and n+ 2 be the integers.
1/n + 1/(n+2) = 12/35
35n(n+2) + 35n
= 12n(n+2)
70n + 70 = 12n² + 24n
12n² - 46n - 70 = 0
6n² - 23n - 35 = 0
(6n+7)(n-5) = 0
n = -7/6 or 5
-7/6 cannot be the integer so will take 5 as the answer.
So the final answers are 5 and 7.
If there is any confusion please leave a comment below.
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