Answer :

Certainly! Let's solve the problem step-by-step.

1. Given Equation:

[tex]\[ \left(a + \frac{1}{a} + 2\right)^2 = 4 \][/tex]

2. Simplifying the Equation:

Let [tex]\( x = a + \frac{1}{a} \)[/tex]. Therefore, the equation can be rewritten as:

[tex]\[ \left(x + 2\right)^2 = 4 \][/tex]

3. Expanding and Solving for [tex]\( x \)[/tex]:

Expand the left side of the equation:

[tex]\[ (x + 2)^2 = x^2 + 4x + 4 \][/tex]

Now set it equal to the right side:

[tex]\[ x^2 + 4x + 4 = 4 \][/tex]

Subtract 4 from both sides:

[tex]\[ x^2 + 4x = 0 \][/tex]

Factor the equation:

[tex]\[ x(x + 4) = 0 \][/tex]

4. Solutions for [tex]\( x \)[/tex]:

The factored equation gives us two solutions:

[tex]\[ x = 0 \quad \text{or} \quad x = -4 \][/tex]

5. Analyzing Each Solution:

- For [tex]\( x = 0 \)[/tex]:

[tex]\[ a + \frac{1}{a} = 0 \][/tex]

Multiply both sides by [tex]\( a \)[/tex]:

[tex]\[ a^2 = -1 \][/tex]

Since [tex]\( a^2 = -1 \)[/tex] is not possible for real numbers (as the square of a real number cannot be negative), [tex]\( x = 0 \)[/tex] is not a viable solution.

- For [tex]\( x = -4 \)[/tex]:

[tex]\[ a + \frac{1}{a} = -4 \][/tex]

Multiply both sides by [tex]\( a \)[/tex]:

[tex]\[ a^2 + 1 = -4a \][/tex]

Rearrange to form a standard quadratic equation:

[tex]\[ a^2 + 4a + 1 = 0 \][/tex]

6. Solving the Quadratic Equation:

Use the quadratic formula [tex]\( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = 1 \)[/tex]:

[tex]\[ a = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \][/tex]

Simplify inside the square root:

[tex]\[ a = \frac{-4 \pm \sqrt{16 - 4}}{2} \][/tex]

[tex]\[ a = \frac{-4 \pm \sqrt{12}}{2} \][/tex]

[tex]\[ a = \frac{-4 \pm 2\sqrt{3}}{2} \][/tex]

Simplify further:

[tex]\[ a = -2 \pm \sqrt{3} \][/tex]

Therefore, the solutions for [tex]\( a \)[/tex] are:

[tex]\[ a_1 = -2 + \sqrt{3} \quad \text{and} \quad a_2 = -2 - \sqrt{3} \][/tex]

7. Finding [tex]\( a^2 + \frac{1}{a^2} \)[/tex]:

We use the identity [tex]\( \left(a + \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} + 2 \)[/tex].

Given [tex]\( a + \frac{1}{a} = -4 \)[/tex] and squaring both sides,

[tex]\[ (-4)^2 = a^2 + \frac{1}{a^2} + 2 \][/tex]

Simplify:

[tex]\[ 16 = a^2 + \frac{1}{a^2} + 2 \][/tex]

Subtract 2 from both sides:

[tex]\[ a^2 + \frac{1}{a^2} = 14 \][/tex]

Therefore, the value of [tex]\( a^2 + \frac{1}{a^2} \)[/tex] is [tex]\( 14 \)[/tex].