Question

The sum of two reciprocals of two consecutive positive integers is 17/72. Write an equation that can be used to find the two integers. What are the integers?

*I found them, I just need it checked*

Answer 1

HI!!

Answer 2

did you write something like
\[\frac{1}{n}+\frac{1}{n+1}=\frac{17}{72}\]?

Answer 3

\[\frac{1}{x}+\frac{1}{(x+1)}=\frac{17}{72}\]

Yes lol

Answer 4

kk you win

Answer 5

I'll skip to the quad form....one sec

Answer 6

\[\color\magenta\heartsuit\]

Answer 7

\[\frac{ 1 }{ x }+\frac{ 1 }{ x+1 }=\frac{ 17 }{ 72 },x \neq 0,x \neq-1\]

Answer 8

just guess you will get it instantly

Answer 9

\[8\times 9=72\\
8+9=17\] that is all

Answer 10

\[x=\frac{ -(-127) \pm \sqrt {(-127)^2-4(17)(-72)} }{ 2(17) }\]

I can't guess...I have to show my work :/

Answer 11

But I got a negative for the second number and I don't know what I did wrong :/

Answer 12

After I did that, (I'm skipping some steps to save time):

\[\frac {127 \pm 145}{34}\]

x=8
x=-9/17

I'm not sure where I went wrong....

Answer 13

a quadratic gives always two values.
x=8 is integer ,hence is correct
so consecutive integers are x,x+1
?

Answer 14

\[\frac{ x+1+x }{ x(x+1) }=\frac{ 17 }{ 72 }\]
\[\frac{ 2x+1 }{ x^2+x }=\frac{ 17 }{ 72 }\]
\(17 x^2+17x=144x+72\)
\(17x^2-127x-72=0\)
\[x=\frac{ 127\pm \sqrt{(-127)^2-4*17*72} }{ 2*17 }\]\[x=\frac{ 127\pm \sqrt{16129+4896} }{ 34 }=\frac{ 127\pm \sqrt{21025} }{ 34 }\]
\[=\frac{ 127\pm145 }{ 34 }\]
so you are correct .

Answer 15

correction -4*17*-72=+4896

Answer 16

Oh! Okay...thank you!