Solution: We are given $ x + \frac{1}{x} = 4 $. Square both sides:
["Solving the Equation: A Powerful Algebraic Technique — From $ x + \frac{1}{x} = 4 $", "When faced with the equation\n$$ x + \frac{1}{x} = 4, $$\na common and highly effective strategy in algebra is to square both sides of the equation. This simple yet powerful manipulation can transform a nonlinear expression into a more solvable form, revealing valuable insights and leading to the final solution.", "---", "### Why Square Both Sides?", "Squaring both sides of $ x + \frac{1}{x} = 4 $ eliminates the fraction and reveals a quadratic structure. It’s particularly useful in solving symmetric expressions involving $ x $ and $ \frac{1}{x} $, and it forms the foundation for solving more complex equations in algebra and beyond.", "---", "### Step-by-Step Solution", "We begin with the given equation:\n$$\nx + \frac{1}{x} = 4\n$$", "#### Step 1: Square both sides", "$$\n\left(x + \frac{1}{x}\right)^2 = 4^2\n$$", "#### Step 2: Expand the left-hand side", "Using the identity $ (a + b)^2 = a^2 + 2ab + b^2 $, we expand:\n$$\nx^2 + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 16\n$$", "Simplify the middle term:\n$$\nx^2 + 2 + \frac{1}{x^2} = 16\n$$", "#### Step 3: Rearranging the equation", "Subtract 2 from both sides:\n$$\nx^2 + \frac{1}{x^2} = 14\n$$", "This new expression can now be used to find $ x $ by treating it as a quadratic in disguise.", "---", "### Solving the Transformed Equation", "We now know:\n$$\nx^2 + \frac{1}{x^2} = 14\n$$", "Multiply both sides by $ x^2 $ (noting $ x <br/>\neq 0 $) to eliminate the denominator:\n$$\nx^4 + 1 = 14x^2\n$$", "Bring all terms to one side:\n$$\nx^4 - 14x^2 + 1 = 0\n$$", "Let $ y = x^2 $, then the equation becomes:\n$$\ny^2 - 14y + 1 = 0\n$$", "Apply the quadratic formula:\n$$\ny = \frac{14 \pm \sqrt{(-14)^2 - 4(1)(1)}}{2} = \frac{14 \pm \sqrt{196 - 4}}{2} = \frac{14 \pm \sqrt{192}}{2}\n$$", "Simplify $ \sqrt{192} = \sqrt{64 \cdot 3} = 8\sqrt{3} $, so:\n$$\ny = \frac{14 \pm 8\sqrt{3}}{2} = 7 \pm 4\sqrt{3}\n$$", "Recall $ y = x^2 $, so:\n$$\nx^2 = 7 + 4\sqrt{3} \quad \ ext{or} \quad x^2 = 7 - 4\sqrt{3}\n$$", "Now take square roots (noting that both values are positive, so real solutions exist):\n$$\nx = \pm \sqrt{7 + 4\sqrt{3}} \quad \ ext{or} \quad x = \pm \sqrt{7 - 4\sqrt{3}}\n$$", "---", "### Final Thoughts", "By squaring $ x + \frac{1}{x} = 4 $, we successfully transformed a symmetric rational expression into a solvable quartic equation. This technique not only simplifies the problem but also reveals elegant algebraic structure—ideal for both math students and professionals solving complex equations.", "If you're working with expressions involving $ x + \frac{1}{x} $, remember: squaring both sides is often the first step toward simplification and solving.", "---", "### Key Takeaways:", "- Squaring $ x + \frac{1}{x} = 4 $ yields $ x^2 + \frac{1}{x^2} = 14 $\n- This leads to a quadratic in $ x^2 $, solvable via the quadratic formula\n- Real solutions are $ x = \pm \sqrt{7 \pm 4\sqrt{3}} $", "Keywords: solve $ x + \frac{1}{x} = 4 $, squaring both sides method, algebraic manipulation, quadratic in reciprocal form, $ x^2 + \frac{1}{x^2} $, real solutions of symmetric equations.", "Meta Description: Learn how squaring $ x + \frac{1}{x} = 4 $ simplifies to $ x^2 + \frac{1}{x^2} = 14 $, leading to exact solutions. Master this algebraic method today."]
