Certainly! To find \( x^2 + \frac{1}{x^2} \) and \( x + \frac{1}{x} \) given that \( x^4 + \frac{1}{x^4} = 119 \), we can proceed with the following steps:
Step 1: Relate \( x^2 + \frac{1}{x^2} \) to the given equation
Let:
[tex]\[ y = x^2 + \frac{1}{x^2} \][/tex]
We know a useful identity:
[tex]\[ \left( x^2 + \frac{1}{x^2} \right)^2 = x^4 + 2 + \frac{1}{x^4} \][/tex]
Rearranging this gives us:
[tex]\[ y^2 = x^4 + 2 + \frac{1}{x^4} \][/tex]
Substituting the given value \( x^4 + \frac{1}{x^4} = 119 \):
[tex]\[ y^2 = 119 + 2 \][/tex]
[tex]\[ y^2 = 121 \][/tex]
Taking the square root of both sides:
[tex]\[ y = \sqrt{121} \][/tex]
[tex]\[ y = \pm 11 \][/tex]
So:
[tex]\[ x^2 + \frac{1}{x^2} = 11 \text{ or } x^2 + \frac{1}{x^2} = -11 \][/tex]
Step 2: Relate \( x + \frac{1}{x} \) to \( x^2 + \frac{1}{x^2} \)
We introduce another variable:
[tex]\[ z = x + \frac{1}{x} \][/tex]
Again, we know another useful identity:
[tex]\[ \left( x + \frac{1}{x} \right)^2 = x^2 + 2 + \frac{1}{x^2} \][/tex]
Rearranging this gives us:
[tex]\[ z^2 = x^2 + 2 + \frac{1}{x^2} \][/tex]
[tex]\[ z^2 = y + 2 \][/tex]
For \( y = 11 \):
[tex]\[ z^2 = 11 + 2 \][/tex]
[tex]\[ z^2 = 13 \][/tex]
Taking the square root:
[tex]\[ z = \sqrt{13} \][/tex]
[tex]\[ z = \pm \sqrt{13} \][/tex]
So:
[tex]\[ x + \frac{1}{x} = \sqrt{13} \text{ or } x + \frac{1}{x} = -\sqrt{13} \][/tex]
For \( y = -11 \):
[tex]\[ z^2 = -11 + 2 \][/tex]
[tex]\[ z^2 = -9 \][/tex]
[tex]\[ z = \pm 3i \][/tex]
So:
[tex]\[ x + \frac{1}{x} = 3i \text{ or } x + \frac{1}{x} = -3i \][/tex]
Summary of Solutions:
1. \( x^2 + \frac{1}{x^2} = 11 \text{ or } -11 \)
2. Corresponding to \( x^2 + \frac{1}{x^2} = 11 \):
[tex]\[ x + \frac{1}{x} = \pm \sqrt{13} \][/tex]
3. Corresponding to \( x^2 + \frac{1}{x^2} = -11 \):
[tex]\[ x + \frac{1}{x} = \pm 3i \][/tex]
Thus, the possible values for [tex]\( x^2 + \frac{1}{x^2} \)[/tex] are [tex]\( 11 \)[/tex] and [tex]\( -11 \)[/tex], and the possible values for [tex]\( x + \frac{1}{x} \)[/tex] are [tex]\( \pm \sqrt{13} \)[/tex] and [tex]\( \pm 3i \)[/tex].
