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The reciprocal part of $x$'s non-integer decimal part equals $x+1$, and $x>0$. What is $x$?

Solution: I tried this way-

Let's $n$= integer part of $x$

$1/x-n = x+1$

or, $1=(x-n)(x+1)$

or, $1= x^2+x-nx-n$

or, $x^2+ (1-n)x -(n+1)=0$

but, stucked here. Is there any other way?

haqnatural
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Mahmudul Hasan
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2 Answers2

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Solve for $x$ using the quadratic formula. As stated, there are infinitely many solutions...

Ashvin Swaminathan
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To complete Ashvin Swaminathan's answer:

$$x = \frac{n-1 \pm \sqrt{n^2 + 2n + 5}}{2}$$

Because $x > 0$, we take $$x = \frac{n-1 + \sqrt{n^2 + 2n + 5}}{2}$$

Since the discriminant $n^2 + 2n + 5 = (n+1)^2 + 4 \geq 4$, then there are indeed infinitely many solutions, and these depend on $n = \lfloor{x}\rfloor$.

Community
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Jose Arnaldo Bebita Dris
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