Nice to meet you, 1/a!

Algebra Level pending

Let $a,b,c$ three distinct nonzero real numbers such that

$a+\frac { 1 }{ b } =b+\frac { 1 }{ c } =c+\frac { 1 }{ a }$

find $\left| abc \right|$

This question is flagged because such a scenario cannot occur, which makes the question pointless.

The answer is 1.

1 solution

Ww Margera
Sep 23, 2014

It is clear that the intended solution is something like this:

$a + 1/a = b + 1/b => a-b = (a-b)/(ab) => ab = 1$ as a and b are not the same.

Similarly ac = 1 and bc = 1. Multiplying these three equations, $(abc)^2=1 => |abc| = 1$ .

Unfortunately, it is not possible to find three distinct a, b and c like this. Here is why:

Suppose $a + 1/a = b + 1/b = c + 1/c = S$ . Then the equation $x + 1/x = S$ has three distinct roots. But this amounts to saying that $x^2 - Sx + 1 = 0$ , which is only a quadratic equation. So it cannot have three distinct roots.

Actually, if there are three roots to a quadratic equation, it implies that it is not an equation, but rather an identity. Thus, there are infinite roots, other than just $a,b,c$

See the solution for this problem

Nanayaranaraknas Vahdam - 6 years, 8 months ago

In the example you gave, you can simplify the lhs to get 1=1. This is different as the coefficient of the quadratic term is known to be one and so the expression cannot be identically 0. This is the same as how you cannot get three distinct numbers to satisfy $x^2-1=0$

ww margera - 6 years, 8 months ago

Nice to meet you, 1/a!

The answer is 1.

1 solution

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