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

Solutions

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

If \(x + \frac{1}{x} = 4\), what is \(\lvert x - \frac{1}{x} \rvert\) ?

We will show two methods to solve the problem.

Method 1

We rewrite the absolute value with square and sqaure root.

\[\lvert x - \frac{1}{x} \rvert = \sqrt{ (x - \frac{1}{x})^2 } = \sqrt{ x^2 - 2 + \frac{1}{x^2} } = \sqrt{ x^2 + \frac{1}{x^2} - 2}.\]

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,

\[\lvert x - \frac{1}{x} \rvert = \sqrt{ x^2 + \frac{1}{x^2} - 2} = \sqrt{ 14 - 2} = \sqrt{12} = 2\sqrt{3}.\]

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

\[x^2 + 1 = 4x\]

Subtracting \(4x\) from both sides, we have

\[x^2 - 4x + 1 = 0\]

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.

\[\lvert x - \frac{1}{x} \rvert = (2 + \sqrt{3}) - \frac{1}{ (2 + \sqrt{3}) } = \frac{ (2 + \sqrt{3})^2 - 1 }{2 + \sqrt{3}} = \frac{ 2^2 + 4\sqrt{3} + 3 - 1}{2 + \sqrt{3}} = \frac{ 6 + 4\sqrt{3}}{2 + \sqrt{3}}\]

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

\[\begin{align} \lvert x - \frac{1}{x} \rvert & = \frac{ 6 + 4\sqrt{3}}{2 + \sqrt{3}} \\ & = \frac{ (6 + 4\sqrt{3})(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})}) \\ & = \frac{ 12 - 6\sqrt{3} + 8\sqrt{3} - 4(\sqrt{3})^2}{2^2 - (\sqrt{3})^2} \\ & = \frac{ 12 + 2\sqrt{3} - 12}{4 - 3} \\ & = \frac{ 2\sqrt{3} }{1} \\ & = 2\sqrt{3} \end{align}\]

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:

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

or

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