Question

The sum of the reciprocals of two consecutive integers is -9/20. find the integers

Answer 1

letthe first integer be x,let the second integer be x+2 so \[\frac{ 1 }{ x }+ \frac{ 1 }{ x+2 }=-\frac{ 9 }{ 20 }\]

Answer 2

\[\frac{ 20(x)(x+2) }{x }+ \frac{ 20(x)(x+2) }{ x+2 }=-\frac{ 20(x)(x+2) *9}{ 20 }\] cancel out all like-terms. \[20(x+2)+20(x)=9x(x+2)\] simplify and solve for x

Answer 3

A reciprocal of x is 1/x. (e.g. reciprocal of 2 is 1/2)

Assuming on of the integer is x, the two consecutive integers would be x and x+1.

Sum of their reciprocals,

\[\frac{ 1 }{ x } + \frac{1}{x+1} = - \frac{9}{20}\]
\[\frac{1 \times (x+1)}{x \times (x+1)} + \frac{1 \times x}{ (x+1) \times x} = - \frac{9}{20}\]
\[\frac{ (x+1) + (x)}{x^2 + x} = - \frac{9}{20}\]
\[\frac{2x+1}{x^2+x} = - \frac{9}{20}\]
Cross multiply them,
\[20 (2x +1) = -9 (x^2 +x) \]
\[40x + 20 = -9x^2 -9x\]
Equate them to 0 to form a quadratic equation,
\[9x^2 +49x +20 = 0\]

Solving for quadratic equation would give you
x = -5 or x = -0.444.

SInce x is an integer, x = -5.

The two integers are -5 and -4.