Answer :
To find the value of [tex]\( x^2 + \frac{1}{x^2} \)[/tex] given that [tex]\( x - \frac{1}{x} = 3 \)[/tex], follow this procedure:
1. Start with the given equation:
[tex]\[ x - \frac{1}{x} = 3 \][/tex]
2. Square both sides of the equation to eliminate the fraction:
[tex]\[ \left( x - \frac{1}{x} \right)^2 = 3^2 \][/tex]
3. Expand the left side using the binomial theorem:
[tex]\[ \left( x - \frac{1}{x} \right)^2 = x^2 - 2 \cdot x \cdot \frac{1}{x} + \left( \frac{1}{x} \right)^2 \][/tex]
[tex]\[ x^2 - 2 + \frac{1}{x^2} = 9 \][/tex]
4. Remember, [tex]\( x \cdot \frac{1}{x} = 1 \)[/tex], so:
[tex]\[ x^2 + \frac{1}{x^2} - 2 = 9 \][/tex]
5. Isolate [tex]\( x^2 + \frac{1}{x^2} \)[/tex] by adding 2 to both sides of the equation:
[tex]\[ x^2 + \frac{1}{x^2} = 9 + 2 \][/tex]
[tex]\[ x^2 + \frac{1}{x^2} = 11 \][/tex]
Thus, the value of [tex]\( x^2 + \frac{1}{x^2} \)[/tex] is [tex]\( \boxed{11} \)[/tex].
1. Start with the given equation:
[tex]\[ x - \frac{1}{x} = 3 \][/tex]
2. Square both sides of the equation to eliminate the fraction:
[tex]\[ \left( x - \frac{1}{x} \right)^2 = 3^2 \][/tex]
3. Expand the left side using the binomial theorem:
[tex]\[ \left( x - \frac{1}{x} \right)^2 = x^2 - 2 \cdot x \cdot \frac{1}{x} + \left( \frac{1}{x} \right)^2 \][/tex]
[tex]\[ x^2 - 2 + \frac{1}{x^2} = 9 \][/tex]
4. Remember, [tex]\( x \cdot \frac{1}{x} = 1 \)[/tex], so:
[tex]\[ x^2 + \frac{1}{x^2} - 2 = 9 \][/tex]
5. Isolate [tex]\( x^2 + \frac{1}{x^2} \)[/tex] by adding 2 to both sides of the equation:
[tex]\[ x^2 + \frac{1}{x^2} = 9 + 2 \][/tex]
[tex]\[ x^2 + \frac{1}{x^2} = 11 \][/tex]
Thus, the value of [tex]\( x^2 + \frac{1}{x^2} \)[/tex] is [tex]\( \boxed{11} \)[/tex].
