An algebra problem by Guilherme Dela Corte

Algebra Level 2

$x + \dfrac{1}{x} + \dfrac{1}{x + \dfrac{1}{x}} = 4$

Which of the following options is true?

The equation has

2

real roots, none of which is

x = 1

The equation only has the root

x = 1

The equation has

2

real roots, one of which is

x = 1

2 solutions

Guilherme Dela Corte
Mar 6, 2017

It can easily be seen that $1 + \frac{1}{1} + \dfrac{1}{1 + \frac{1}{1}} = 2.5 \neq 4$ .

Let $x + \dfrac{1}{x} = s$ . We can see that $s + \dfrac{1}{s} = 4$ has roots $s = 2 \pm \sqrt{3}$ .

Since $\left | x + \dfrac{1}{x} \right | \geq 2, \; \forall x \in \mathbb{R}$ , we have that $s = x + \dfrac{1}{x} = 2 -\sqrt{3}$ yields no real roots.

However, $s = x + \dfrac{1}{x} = 2 + \sqrt{3}$ yields two distinct real roots.

Hana Wehbi
Mar 7, 2017

By plugging $1$ into the equation, we see it doesn't satisfy the equation, thus it is not a root.

You have only shown that $x=1$ cannot be the solution. How do you know that the equation has 2 real roots?

Pi Han Goh - 4 years, 3 months ago

@Pi Han Goh , Yes, you are right. I also agree with the author solution, but what I was trying to say that since we only have one answer that says $x\ne 1$ , we can easily see that from the equation.

Hana Wehbi - 4 years, 3 months ago

Then you didn't show that you properly solved this question, where you have only eliminated the wrong options, but didn't justify the the remaining option is indeed correct.

Pi Han Goh - 4 years, 3 months ago

@Pi Han Goh , No definitely my statement is not a solution, just a hint of tackling the problem.

Hana Wehbi - 4 years, 3 months ago

An algebra problem by Guilherme Dela Corte

2 solutions

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