Sum of reciprocals of 2 consecutive integers is 7/24. What are the 2 integers?

Answer 1

1/n+1/(n+1) =7/24
(2n+1)/(n^2+n)= 7/24
24(2n+1) =7(n^2+n)

7n^2 -41n-24=0
n=((41+-sqrt(1681+672))/14
but 1681+672 is not a perfect square so there is no solution

Answer 2

think first of two even consecutive integers is 'x'. for this reason, next consecutive integer is 'x+2'. [e.g. if between the first consecutive integers is '6', then the subsequent even consecutive integer must be '8']. Sum of their reciprocals is 7/24. for this reason, a million/x + a million/(x+2) = 7/24 Simplifying, ((x+2) + x) / (x*(x+2)) = 7/24 more advantageous simplifying, (2x + 2)*24 = 7x(x+2) this offers us 48x+40 8 = 7x^2 + 14x which turn outs to be, 7x^2 - 34x - 40 8 = 0 to locate 'x' out of this equation, we've locate 2 aspects such that their sum is -34 and their multiplication will outcome in 7 * (-40 8) = -336. Such 2 aspects are -40 2 and +8. [Their sum is -34 and their multiplication is -336.] for that reason, our equation turns to be 7x^2 - 42x + 8x - 40 8 = 0 this offers us, 7x(x-6) + 8(x-6) = 0 that's, (x-6)(7x+8) = 0 for this reason, 'x' can both be '6' or it would properly be '-8/7'. As 'x' is an integer, answer is '6'. for that reason 2 consecutive integers are '6' and '8'. [x=6, x+2=6+2=8] you could ensure the answer: a million/6 + a million/8 = 14/40 8 = 7/24. for this reason, answer is shown.

Answer 3

let the consecutive integers be x , x+1
1/x + 1/ x+1 = 7/24

(x +1 ) + x / x(x+1 ) = 24( x + x +1 ) = 7 (x^2 + x )

24(2x +1) = 7x^2 +7x

48x +24 = 7x^2 + 7x

7x^2 - 41 x - 24 =0

x cannot be an integer.
thankyou

7x^2 - 41 x - 24

Answer 4

1/x + 1/(x+1) = 7/24

(x+1) + x = (7/24)(x)(x+1)

2x + 1 = (7/24)(x)(x+1)

0 = (7/24)x^2 -2x -1+ (7/24)

0 = (7/24)x^2 -2x -17/24

0 = 7x^2 - 48x - 17

x = 7.1947