I’ll walk you through this solution step by step and provide an explanation.

Problem Statement

Given: $a + \frac{1}{a} = 2$ we need to prove: $a^2 + \frac{1}{a^2} = 2 \quad \text{and} \quad a^4 + \frac{1}{a^4} = 2$

Step-by-Step Solution

Step 1: Solving for $a^2 + \frac{1}{a^2}$

1. Starting with the given equation: $a + \frac{1}{a} = 2$

2. Square both sides to find $a^2 + \frac{1}{a^2}$: $\left( a + \frac{1}{a} \right)^2 = 2^2$

3. Expanding the square: $a^2 + 2 \cdot a \cdot \frac{1}{a} + \frac{1}{a^2} = 4$ Here, $2 \cdot a \cdot \frac{1}{a} = 2$, so we simplify to: $a^2 + \frac{1}{a^2} + 2 = 4$

4. Solving for $a^2 + \frac{1}{a^2}$: $a^2 + \frac{1}{a^2} = 4 - 2 = 2$

So, we’ve shown that: $a^2 + \frac{1}{a^2} = 2$

Step 2: Solving for $a^4 + \frac{1}{a^4}$

1. Square both sides of the previous result to find $a^4 + \frac{1}{a^4}$: $\left( a^2 + \frac{1}{a^2} \right)^2 = 2^2$

2. Expanding the square: $a^4 + 2 \cdot a^2 \cdot \frac{1}{a^2} + \frac{1}{a^4} = 4$ Here, $2 \cdot a^2 \cdot \frac{1}{a^2} = 2$, so we simplify to: $a^4 + \frac{1}{a^4} + 2 = 4$

3. Solving for $a^4 + \frac{1}{a^4}$: $a^4 + \frac{1}{a^4} = 4 - 2 = 2$

Thus, we’ve shown that: $a^4 + \frac{1}{a^4} = 2$

Conclusion

Both parts are proved:

1. $a^2 + \frac{1}{a^2} = 2$
2. $a^4 + \frac{1}{a^4} = 2$

I’ll create a video based on this explanation to illustrate each step visually. Meanwhile, let me know if you have any questions or if you want additional details about any step.

Relative Questions:

1. Why is it necessary to square both sides of the equation?
2. Can this technique be applied to other values of $a + \frac{1}{a}$?
3. How would the solution change if $a + \frac{1}{a}$ was not equal to 2?
4. What if we wanted to find $a^6 + \frac{1}{a^6}$?
5. How does this method leverage symmetry in algebraic expressions?

Tip:

When solving these kinds of problems, squaring both sides of an equation is a common technique to reach higher powers, but always remember to simplify carefully after expanding.