hi mates pls can you help me
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since the no are odd consecutive
so one no. be x and other be x+2
now according to question
1/x + 1/(x+2)= 12/35
(2x+2 )35 =( x^2 +2x)12
70x + 70 = 24x^2 +24x
24x^2 -47x -70 = 0
plz apply formulae
(-b +-√{b^2-4ac} )/2a
ty
so one no. be x and other be x+2
now according to question
1/x + 1/(x+2)= 12/35
(2x+2 )35 =( x^2 +2x)12
70x + 70 = 24x^2 +24x
24x^2 -47x -70 = 0
plz apply formulae
(-b +-√{b^2-4ac} )/2a
ty
rehan2282:
iska answer nahi ara matlab root nahi nikalraha hai bhai
Answered by
1
Sure I will help you.
Let the first small odd number be a.
Then the next odd be a+2.
a+1 will be an even number.
Their reciprocal's sum is 12/35
Reciprocal is 1/a and 1/a+2
So basically:
1/a+1/a+2=12/35
Multiply the denominators:
(a+2+a)/a²+2a=12/35
2a+2=12a²+24a/35
70a+70=12a²+24a
=12a²-46a-70=0
=12a²-60a+14a-70=0
12a(a-5)+14(a-5)=0
(12a+14)(a-5)=0
a=5 or a=-14/12
but a is natural so a=5.
a+2=5+2=7
The 2 numbers are 5 and 7
Hope it helps you.
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