Answer : The two consecutive integers are, 4 and 5
Step-by-step explanation :
Let the two consecutive number be, x and (x+1)
The sum of the reciprocals of two consecutive integers is 9/20.
The expression will be:
[tex]\frac{1}{x}+\frac{1}{x+1}=\frac{9}{20}[/tex]
Now solving the term 'x', we get:
[tex]\frac{x+1+x}{(x)(x+1)}=\frac{9}{20}[/tex]
[tex]\frac{2x+1}{x^2+x}=\frac{9}{20}[/tex]
[tex]20(2x+1)=9(x^2+x)[/tex]
[tex]40x+20=9x^2+9x[/tex]
[tex]9x^2-31x-20=0[/tex]
We are solving the quadratic equation by middle term splitting.
[tex]9x^2-36x-5x-20=0[/tex]
[tex]9x(x-4)+5(x-4)=0[/tex]
[tex](9x+5)(x-4)=0[/tex]
[tex](9x+5)=0\text{ and }(x-4)=0[/tex]
[tex]x=\frac{-5}{9}\text{ and }x=4[/tex]
We are neglecting the value of [tex]x=\frac{-5}{9}[/tex]
Thus we are taking x = 4
When x = 4 then (x+1) = (4+1) = 5
Thus, the two consecutive integers are, 4 and 5
