Answer :
To solve the problem [tex]\(x = 3 + 2 \sqrt{2}\)[/tex] and to find the value of [tex]\(x^2 + \frac{1}{x^2}\)[/tex], we’ll go through a step-by-step solution:
1. Determine [tex]\(x^2\)[/tex]:
First, we need to compute [tex]\(x^2\)[/tex]:
[tex]\[ x = 3 + 2\sqrt{2} \][/tex]
[tex]\[ x^2 = (3 + 2\sqrt{2})^2 \][/tex]
Expanding the square:
[tex]\[ x^2 = (3)^2 + 2 \cdot 3 \cdot 2\sqrt{2} + (2\sqrt{2})^2 \][/tex]
Simplifying each term:
[tex]\[ x^2 = 9 + 12\sqrt{2} + 8 \][/tex]
[tex]\[ x^2 = 17 + 12\sqrt{2} \][/tex]
2. Calculate [tex]\(\frac{1}{x}\)[/tex]:
To find [tex]\(\frac{1}{x}\)[/tex], notice we need:
[tex]\[ \frac{1}{x^2} \][/tex]
First calculate [tex]\(\frac{1}{x}\)[/tex] by rationalizing:
[tex]\[ \frac{1}{3 + 2\sqrt{2}} \][/tex]
Multiply numerator and denominator by the conjugate:
[tex]\[ \frac{1}{3 + 2\sqrt{2}} \cdot \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \][/tex]
[tex]\[ = \frac{3 - 2 \sqrt{2}}{(3 + 2 \sqrt{2})(3 - 2 \sqrt{2})} \][/tex]
[tex]\[ = \frac{3 - 2 \sqrt{2}}{9 - 8} = 3 - 2 \sqrt{2} \][/tex]
Next, compute the square:
[tex]\[ \left(\frac{1}{x}\right)^2 = (3-2\sqrt{2})^2 \][/tex]
[tex]\[ \left(\frac{1}{x}\right)^2 = 9 - 12\sqrt{2} + 8 \][/tex]
[tex]\[ \left(\frac{1}{x}\right)^2 = 17 - 12\sqrt{2} \][/tex]
3. Sum [tex]\(x^2\)[/tex] and [tex]\(\frac{1}{x^2}\)[/tex]:
Now sum [tex]\(x^2\)[/tex] and [tex]\(\frac{1}{x^2}\)[/tex]:
[tex]\[ x^2 + \frac{1}{x^2} = (17 + 12\sqrt{2}) + (17 - 12\sqrt{2}) \][/tex]
[tex]\[ x^2 + \frac{1}{x^2} = 17 + 17 \][/tex]
[tex]\[ x^2 + \frac{1}{x^2} = 34 \][/tex]
Thus, the value of [tex]\(x^2 + \frac{1}{x^2}\)[/tex] is [tex]\(\boxed{34}\)[/tex].
1. Determine [tex]\(x^2\)[/tex]:
First, we need to compute [tex]\(x^2\)[/tex]:
[tex]\[ x = 3 + 2\sqrt{2} \][/tex]
[tex]\[ x^2 = (3 + 2\sqrt{2})^2 \][/tex]
Expanding the square:
[tex]\[ x^2 = (3)^2 + 2 \cdot 3 \cdot 2\sqrt{2} + (2\sqrt{2})^2 \][/tex]
Simplifying each term:
[tex]\[ x^2 = 9 + 12\sqrt{2} + 8 \][/tex]
[tex]\[ x^2 = 17 + 12\sqrt{2} \][/tex]
2. Calculate [tex]\(\frac{1}{x}\)[/tex]:
To find [tex]\(\frac{1}{x}\)[/tex], notice we need:
[tex]\[ \frac{1}{x^2} \][/tex]
First calculate [tex]\(\frac{1}{x}\)[/tex] by rationalizing:
[tex]\[ \frac{1}{3 + 2\sqrt{2}} \][/tex]
Multiply numerator and denominator by the conjugate:
[tex]\[ \frac{1}{3 + 2\sqrt{2}} \cdot \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \][/tex]
[tex]\[ = \frac{3 - 2 \sqrt{2}}{(3 + 2 \sqrt{2})(3 - 2 \sqrt{2})} \][/tex]
[tex]\[ = \frac{3 - 2 \sqrt{2}}{9 - 8} = 3 - 2 \sqrt{2} \][/tex]
Next, compute the square:
[tex]\[ \left(\frac{1}{x}\right)^2 = (3-2\sqrt{2})^2 \][/tex]
[tex]\[ \left(\frac{1}{x}\right)^2 = 9 - 12\sqrt{2} + 8 \][/tex]
[tex]\[ \left(\frac{1}{x}\right)^2 = 17 - 12\sqrt{2} \][/tex]
3. Sum [tex]\(x^2\)[/tex] and [tex]\(\frac{1}{x^2}\)[/tex]:
Now sum [tex]\(x^2\)[/tex] and [tex]\(\frac{1}{x^2}\)[/tex]:
[tex]\[ x^2 + \frac{1}{x^2} = (17 + 12\sqrt{2}) + (17 - 12\sqrt{2}) \][/tex]
[tex]\[ x^2 + \frac{1}{x^2} = 17 + 17 \][/tex]
[tex]\[ x^2 + \frac{1}{x^2} = 34 \][/tex]
Thus, the value of [tex]\(x^2 + \frac{1}{x^2}\)[/tex] is [tex]\(\boxed{34}\)[/tex].
