IDNLearn.com: Your trusted platform for finding reliable answers. Ask any question and receive accurate, in-depth responses from our dedicated team of experts.
Sagot :
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].
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. For clear and precise answers, choose IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.