Answer :
Given [tex]\( x = 3 - \sqrt{32} \)[/tex], we need to find the value of [tex]\( x^2 + \frac{1}{x^2} \)[/tex].
1. Calculate [tex]\( x^2 \)[/tex]:
[tex]\[ x = 3 - \sqrt{32} \] \[ x^2 = (3 - \sqrt{32})^2 \] \[ x^2 = 3^2 - 2 \cdot 3 \cdot \sqrt{32} + (\sqrt{32})^2 \] \[ x^2 = 9 - 6\sqrt{32} + 32 \] \[ x^2 = 41 - 6\sqrt{32} \][/tex]
2. Calculate [tex]\( \frac{1}{x} \)[/tex]:
[tex]\[ \frac{1}{x} = \frac{1}{3 - \sqrt{32}} \][/tex]
To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator:
[tex]\[ \frac{1}{x} = \frac{1}{3 - \sqrt{32}} \cdot \frac{3 + \sqrt{32}}{3 + \sqrt{32}} \] \[ \frac{1}{x} = \frac{3 + \sqrt{32}}{(3 - \sqrt{32})(3 + \sqrt{32})} \] \[ \frac{1}{x} = \frac{3 + \sqrt{32}}{3^2 - (\sqrt{32})^2} \] \[ \frac{1}{x} = \frac{3 + \sqrt{32}}{9 - 32} \] \[ \frac{1}{x} = \frac{3 + \sqrt{32}}{-23} \] \[ \frac{1}{x} = -\frac{3 + \sqrt{32}}{23} \][/tex]
3. Calculate [tex]\( \left(\frac{1}{x}\right)^2 \)[/tex]:
[tex]\[ \left(\frac{1}{x}\right)^2 = \left(-\frac{3 + \sqrt{32}}{23}\right)^2 \] \[ \left(\frac{1}{x}\right)^2 = \frac{(3 + \sqrt{32})^2}{23^2} \] \[ \left(\frac{1}{x}\right)^2 = \frac{3^2 + 2 \cdot 3 \cdot \sqrt{32} + (\sqrt{32})^2}{529} \] \[ \left(\frac{1}{x}\right)^2 = \frac{9 + 6\sqrt{32} + 32}{529} \] \[ \left(\frac{1}{x}\right)^2 = \frac{41 + 6\sqrt{32}}{529} \][/tex]
4. Combine [tex]\( x^2 \)[/tex] and [tex]\( \left(\frac{1}{x}\right)^2 \)[/tex]:
[tex]\[ x^2 + \frac{1}{x^2} = (41 - 6\sqrt{32}) + \frac{41 + 6\sqrt{32}}{529} \] \[ x^2 + \frac{1}{x^2} = 41 - 6\sqrt{32} + \frac{41}{529} + \frac{6\sqrt{32}}{529} \][/tex]
Since the term ([tex]\( - 6\sqrt{32} \)[/tex] ) and ( [tex]+ 6\sqrt{32}[/tex] ) cancel each other out, we only need to add the constants:
[tex]\[ x^2 + \frac{1}{x^2} = 41 + \frac{41}{529} \] \[ x^2 + \frac{1}{x^2} = 41 + \frac{41}{529} \] \[ x^2 + \frac{1}{x^2} = 41 + 0.07751 \approx 41.0775 \][/tex]
So the value of [tex]\( x^2 + \frac{1}{x^2} \)[/tex] is approximately [tex]\( 41.0775 \)[/tex].
