Answer :
Sure, let's work through this step by step:
Given:
[tex]\[ P = x - \frac{1}{x} \][/tex]
and
[tex]\[ q = x^2 + \frac{1}{x^2} \][/tex]
We want to express [tex]\( q \)[/tex] in terms of [tex]\( P \)[/tex].
1. Start by squaring both sides of the equation [tex]\( P = x - \frac{1}{x} \)[/tex]:
[tex]\[ \left( x - \frac{1}{x} \right)^2 = P^2 \][/tex]
2. Expand the left-hand side:
[tex]\[ \left( x - \frac{1}{x} \right)^2 = x^2 - 2 \cdot x \cdot \frac{1}{x} + \left( \frac{1}{x} \right)^2 \][/tex]
3. Simplify the expression:
[tex]\[ x^2 - 2 + \frac{1}{x^2} = P^2 \][/tex]
4. Rearrange the equation to solve for [tex]\( x^2 + \frac{1}{x^2} \)[/tex]:
[tex]\[ x^2 + \frac{1}{x^2} = P^2 + 2 \][/tex]
Thus, [tex]\( q \)[/tex] in terms of [tex]\( P \)[/tex] is:
[tex]\[ q = P^2 + 2 \][/tex]
So, the final expression is:
[tex]\[ q = P^2 + 2 \][/tex]
Given:
[tex]\[ P = x - \frac{1}{x} \][/tex]
and
[tex]\[ q = x^2 + \frac{1}{x^2} \][/tex]
We want to express [tex]\( q \)[/tex] in terms of [tex]\( P \)[/tex].
1. Start by squaring both sides of the equation [tex]\( P = x - \frac{1}{x} \)[/tex]:
[tex]\[ \left( x - \frac{1}{x} \right)^2 = P^2 \][/tex]
2. Expand the left-hand side:
[tex]\[ \left( x - \frac{1}{x} \right)^2 = x^2 - 2 \cdot x \cdot \frac{1}{x} + \left( \frac{1}{x} \right)^2 \][/tex]
3. Simplify the expression:
[tex]\[ x^2 - 2 + \frac{1}{x^2} = P^2 \][/tex]
4. Rearrange the equation to solve for [tex]\( x^2 + \frac{1}{x^2} \)[/tex]:
[tex]\[ x^2 + \frac{1}{x^2} = P^2 + 2 \][/tex]
Thus, [tex]\( q \)[/tex] in terms of [tex]\( P \)[/tex] is:
[tex]\[ q = P^2 + 2 \][/tex]
So, the final expression is:
[tex]\[ q = P^2 + 2 \][/tex]
