Let there be a quotient of two natural numbers in which denominator is one less than square of numerator. If we add 2 to both num. and den. the quotient will excess 2/3. If we subtract 3 from num. as well as den. the quotient will lie between 0 and 1/10. Determine the quotient.

Question

Arun Kumar · Answer

if we try to solve it

n/d=q

d=n^2-1

(n+2)/(n^2+1)=q+2/3

q=n/(n^2-1)

we get a equation

(n+2)/(n^2+1)=n/(n^2-1)+2/3

we get

2n^4-3n^3-3n^2+6n+4=0

one possible solution n=-1

something missing in the question

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Let there be a quotient of two natural numbers in which denominator is one less than square of numerator. If we add 2 to both num. and den. the quotient will excess 2/3. If we subtract 3 from num. as well as den. the quotient will lie between 0 and 1/10. Determine the quotient.

Let there be a quotient of two natural numbers in which denominator is one less than square of numerator. If we add 2 to both num. and den. the quotient will excess 2/3. If we subtract 3 from num. as well as den. the quotient will lie between 0 and 1/10. Determine the quotient.

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