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].
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].
