Solving for \(x^2 - \frac{1}{x^2}\) Given \(x^2 + \frac{1}{x^2}\)

The problem asks us to find the value of \(x^2 - \frac{1}{x^2}\) given that \(x^2 + \frac{1}{x^2} = 7\) and the condition \(0 < x < 1\). This is a classic algebra problem that involves using algebraic identities.

Using Algebraic Identities

We know the identity that relates the square of a difference to the square of a sum:

\((a-b)^2 = (a+b)^2 - 4ab\)

In this problem, we can consider \(a = x^2\) and \(b = \frac{1}{x^2}\). Substituting these into the identity:

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

Notice that the term \(4\left(x^2\right)\left(\frac{1}{x^2}\right)\) simplifies nicely because \(x^2 \times \frac{1}{x^2} = 1\). So the equation becomes:

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

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

Substituting the Given Value

We are given that \(x^2 + \frac{1}{x^2} = 7\). We can substitute this value into the equation we derived:

\(\left(x^2 - \frac{1}{x^2}\right)^2 = (7)^2 - 4\)

\(\left(x^2 - \frac{1}{x^2}\right)^2 = 49 - 4\)

\(\left(x^2 - \frac{1}{x^2}\right)^2 = 45\)

Finding the Value of \(x^2 - \frac{1}{x^2}\)

To find the value of \(x^2 - \frac{1}{x^2}\), we take the square root of both sides:

\(x^2 - \frac{1}{x^2} = \pm \sqrt{45}\)

We can simplify the square root of 45:

\(\sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \times \sqrt{5} = 3\sqrt{5}\)

So, we have two possible values for \(x^2 - \frac{1}{x^2}\):

• \(x^2 - \frac{1}{x^2} = 3\sqrt{5}\)
• \(x^2 - \frac{1}{x^2} = -3\sqrt{5}\)

Using the Condition \(0 < x < 1\)

The problem provides an important condition: \(0 < x < 1\). This condition helps us determine which of the two possible values is correct.

If \(0 < x < 1\), then when we square \(x\), \(x^2\) will also be between 0 and 1 (\(0 < x^2 < 1\)).

Now consider the term \(\frac{1}{x^2}\). If \(x^2\) is a positive number less than 1, then its reciprocal, \(\frac{1}{x^2}\), will be greater than 1.

For example, if \(x = 0.5\), then \(x^2 = (0.5)^2 = 0.25\). Here \(0 < 0.25 < 1\).

The reciprocal is \(\frac{1}{x^2} = \frac{1}{0.25} = 4\). Here \(4 > 1\).

So, we have \(x^2\) which is less than 1, and \(\frac{1}{x^2}\) which is greater than 1. When we subtract a larger number from a smaller positive number, the result is negative.

\(x^2 - \frac{1}{x^2} = (\text{a value between 0 and 1}) - (\text{a value greater than 1})\)

Therefore, \(x^2 - \frac{1}{x^2}\) must be negative.

Conclusion

Since \(x^2 - \frac{1}{x^2}\) must be negative based on the condition \(0 < x < 1\), the correct value from the two possibilities (\(3\sqrt{5}\) and \(-3\sqrt{5}\)) is \(-3\sqrt{5}\).

The value of \(x^2 - \frac{1}{x^2}\) is \(-3\sqrt{5}\).

Revision Table: Understanding \(x^2 \pm \frac{1}{x^2}\)

Problems involving expressions like \(x + \frac{1}{x}\) or \(x^2 + \frac{1}{x^2}\) are common. Here's a quick summary of useful relationships:

• \(\left(x + \frac{1}{x}\right)^2 = x^2 + 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}\)
• So, \(x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2\)
• \(\left(x - \frac{1}{x}\right)^2 = x^2 - 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = x^2 - 2 + \frac{1}{x^2}\)
• So, \(x^2 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 2\)
• From these, we can also see: \(\left(x + \frac{1}{x}\right)^2 - \left(x - \frac{1}{x}\right)^2 = (x^2 + 2 + \frac{1}{x^2}) - (x^2 - 2 + \frac{1}{x^2}) = 4\).
• This gives us the identity \((x - \frac{1}{x})^2 = (x + \frac{1}{x})^2 - 4\), which is a specific case of the identity we used in the solution.

Additional Information: Significance of the Domain

The condition \(0 < x < 1\) is crucial. Without it, \(x^2 - \frac{1}{x^2}\) could be either \(3\sqrt{5}\) or \(-3\sqrt{5}\). The domain restricts the possible values of x, which in turn restricts the possible values of the expression we are trying to find.

Consider the function \(f(x) = x^2 - \frac{1}{x^2}\).

• If \(x > 1\), then \(x^2 > 1\) and \(0 < \frac{1}{x^2} < 1\). In this case, \(x^2 - \frac{1}{x^2}\) would be positive. For example, if \(x=2\), \(x^2=4\), \(\frac{1}{x^2}=\frac{1}{4}\), and \(x^2 - \frac{1}{x^2} = 4 - \frac{1}{4} = 3.75\).
• If \(0 < x < 1\), then \(0 < x^2 < 1\) and \(\frac{1}{x^2} > 1\). In this case, \(x^2 - \frac{1}{x^2}\) would be negative. For example, if \(x=0.5\), \(x^2=0.25\), \(\frac{1}{x^2}=4\), and \(x^2 - \frac{1}{x^2} = 0.25 - 4 = -3.75\).

Thus, the condition \(0 < x < 1\) correctly indicates that the value of \(x^2 - \frac{1}{x^2}\) must be negative.