A Number and Its Reciprocal

If a number $x$ plus its reciprocal is equal to 4, what is the absolute difference between $x$ and its reciprocal? Express the answer in simplest radical form.

Solutions

First, we rewrite the problem with math equations. The question is

If $x + \frac{1}{x} = 4$ , what is $| x - \frac{1}{x} |$ ?

We will show two methods to solve the problem.

Method 1

We rewrite the absolute value with square and sqaure root.

| x - \frac{1}{x} | = \sqrt{(x - \frac{1}{x})^{2}} = \sqrt{x^{2} - 2 + \frac{1}{x^{2}}} = \sqrt{x^{2} + \frac{1}{x^{2}} - 2} .

Next, we find the value of $x^{2} + \frac{1}{x^{2}}$ from the given equation.

Starting from $x + \frac{1}{x} = 4$ , we square both sides of the equation and expand the sqaure of the binomial.

x + \frac{1}{x} = 4

Squaring both sides, we get

(x + \frac{1}{x})^{2} = 4^{2}

Expanding the left side, we get

x^{2} + 2 + \frac{1}{x^{2}} = 16

Subtracting 2 from both sides, we have

x^{2} + \frac{1}{x^{2}} = 14.

Therefore,

| x - \frac{1}{x} | = \sqrt{x^{2} + \frac{1}{x^{2}} - 2} = \sqrt{14 - 2} = \sqrt{12} = 2 \sqrt{3} .

So the answer is $2 \sqrt{3}$ .

Method 2

Since $x + \frac{1}{x} = 4$ , we can try to solve the equation and find out $x$ first.

Multiplying both sides by $x$ , we have

x^{2} + 1 = 4 x

Subtracting $4 x$ from both sides, we have

x^{2} - 4 x + 1 = 0

Solving for $x$ , we have $x = 2 + \sqrt{3}$ or $x = 2 - \sqrt{3} .$

Both values of $x$ should lead to the same answer. Let us try $2 + \sqrt{3}$ first.

| x - \frac{1}{x} | = (2 + \sqrt{3}) - \frac{1}{(2 + \sqrt{3})} = \frac{(2 + \sqrt{3})^{2} - 1}{2 + \sqrt{3}} = \frac{2^{2} + 4 \sqrt{3} + 3 - 1}{2 + \sqrt{3}} = \frac{6 + 4 \sqrt{3}}{2 + \sqrt{3}}

The difficulty here is to remove the square root at the bottom. Below is one way to do it.

\begin{aligned} | x - \frac{1}{x} | & = \frac{6 + 4 \sqrt{3}}{2 + \sqrt{3}} \\ = \frac{(6 + 4 \sqrt{3}) (2 - \sqrt{3})}{(2 + \sqrt{3}) (2 - \sqrt{3})}) \\ = \frac{12 - 6 \sqrt{3} + 8 \sqrt{3} - 4 (\sqrt{3})^{2}}{2^{2} - (\sqrt{3})^{2}} \\ = \frac{12 + 2 \sqrt{3} - 12}{4 - 3} \\ = \frac{2 \sqrt{3}}{1} \\ = 2 \sqrt{3} \end{aligned}

If you try $x = 2 - \sqrt{3}$ , you will get $2 \sqrt{3}$ , too.

Discussions

In method 1, the key is to notice that both the square of $x + \frac{1}{x}$ and the square of $x - \frac{1}{x}$ have $x^{2} + \frac{1}{x^{2}}$ . BTW, the absolute difference of the two squares is always 4.

(x + \frac{1}{x})^{2} - (x - \frac{1}{x})^{2} = (x^{2} + 2 + \frac{1}{x^{2}}) - (x^{2} - 2 + \frac{1}{x^{2}}) = 4.

In method 2, the key is to remove the square root from the denominator. Also, you may have noticed that the reciprocal of $2 + \sqrt{3}$ is $2 - \sqrt{3}$ because their prodcut is 1.

(2 + \sqrt{3}) (2 - \sqrt{3}) = 2^{2} - {\sqrt{3}}^{2} = 4 - 3 = 1.

It may be easy to see in the following equations:

2 + \sqrt{3} = \frac{1}{2 - \sqrt{3}},

2 - \sqrt{3} = \frac{1}{2 + \sqrt{3}} .