To find the value of [tex]\( x^2 + \frac{1}{x^2} \)[/tex] given the equation [tex]\( x - \frac{1}{x} = 2 \)[/tex], follow these steps.
1. Given the equation:
[tex]\[
x - \frac{1}{x} = 2
\][/tex]
2. Isolate terms involving [tex]\( x \)[/tex]:
Multiply both sides of the equation by [tex]\( x \)[/tex] to eliminate the fraction:
[tex]\[
x^2 - 1 = 2x
\][/tex]
Rearrange this to form a standard quadratic equation:
[tex]\[
x^2 - 2x - 1 = 0
\][/tex]
3. Solve the quadratic equation:
Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -1 \)[/tex]:
[tex]\[
x = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}
\][/tex]
So, the solutions are:
[tex]\[
x = 1 + \sqrt{2} \quad \text{or} \quad x = 1 - \sqrt{2}
\][/tex]
4. Calculate [tex]\( x^2 + \frac{1}{x^2} \)[/tex] for each solution:
Let's consider both solutions separately.
- For [tex]\( x = 1 + \sqrt{2} \)[/tex]:
[tex]\[
x^2 = (1 + \sqrt{2})^2 = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2}
\][/tex]
[tex]\[
\frac{1}{x} = \frac{1}{1 + \sqrt{2}} \cdot \frac{1 - \sqrt{2}}{1 - \sqrt{2}} = \frac{1 - \sqrt{2}}{-1} = \sqrt{2} - 1
\][/tex]
[tex]\[
\frac{1}{x}^2 = (\sqrt{2} - 1)^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2}
\][/tex]
Hence,
[tex]\[
x^2 + \frac{1}{x^2} = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6
\][/tex]
- For [tex]\( x = 1 - \sqrt{2} \)[/tex]:
[tex]\[
x^2 = (1 - \sqrt{2})^2 = 1 - 2\sqrt{2} + 2 = 3 - 2\sqrt{2}
\][/tex]
[tex]\[
\frac{1}{x} = \frac{1}{1 - \sqrt{2}} \cdot \frac{1 + \sqrt{2}}{1 + \sqrt{2}} = \frac{1 + \sqrt{2}}{-1} = -(1 + \sqrt{2})
\][/tex]
[tex]\[
\frac{1}{x}^2 = (-(1 + \sqrt{2}))^2 = (1 + \sqrt{2})^2 = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2}
\][/tex]
Hence,
[tex]\[
x^2 + \frac{1}{x^2} = (3 - 2\sqrt{2}) + (3 + 2\sqrt{2}) = 6
\][/tex]
Therefore, in both cases, the value of [tex]\( x^2 + \frac{1}{x^2} \)[/tex] is:
[tex]\[
\boxed{6}
\][/tex]
