Solve: $ b = 12/a $, substitute: $ a^2 - (144/a^2) = -64 $. Multiply by $ a^2 $: $ a^4 + 64a^2 - 144 = 0 $. Let $ u = a^2 $: $ u^2 + 64u - 144 = 0 $.

["Solving the Equation $ b = \frac{12}{a} $: A Step-by-Step Algebraic Transformation", "When tackling equations that begin with simple substitutions, a powerful strategy involves eliminating fractions and simplifying the problem into a solvable polynomial form. In this article, we explore how to solve the equation derived from $ b = \frac{12}{a} $ after substitution, leading to a quadratic in terms of $ a^2 $, and how smart variable substitution transforms the problem into a manageable quadratic.", "---", "### The Original Equation", "We start with:", "[\nb = \frac{12}{a}\n]", "Our goal is to eliminate the denominator by multiplying both sides by $ a $, but first, we proceed as instructed: substitute to simplify.", "Multiply both sides by $ a $:", "[\na b = 12\n]", "Now solve for $ a $:", "[\na = \frac{12}{b}\n]", "This substitution shows a direct inverse relationship, but it’s useful to later manipulate $ a $ algebraically in a transformed equation.", "Instead, let’s follow the detailed algebraic path as originally inspired:", "We are told to substitute $ a^2 - \frac{144}{a^2} = -64 $. Rearranging:", "[\na^2 - \frac{144}{a^2} = -64\n]", "Multiply both sides by $ a^2 $ to eliminate denominators:", "[\na^2 \cdot a^2 - 144 = -64a^2\n]", "[\na^4 - 144 = -64a^2\n]", "Bring all terms to one side:", "[\na^4 + 64a^2 - 144 = 0\n]", "Now, to simplify, let’s make the substitution:", "[\nu = a^2\n]", "Then the equation becomes:", "[\nu^2 + 64u - 144 = 0\n]", "---", "### Solve the Quadratic in $ u $", "We solve $ u^2 + 64u - 144 = 0 $ using the quadratic formula:", "[\nu = \frac{-64 \pm \sqrt{64^2 - 4(1)(-144)}}{2}\n]", "Calculate discriminant:", "[\n64^2 = 4096,\quad 4 \cdot 144 = 576,\quad \ ext{so } \Delta = 4096 + 576 = 4672\n]", "Factor $ 4672 $:", "[\n4672 = 16 \cdot 292 = 16 \cdot 4 \cdot 73 = 64 \cdot 73\n]", "So,", "[\nu = \frac{-64 \pm \sqrt{64 \cdot 73}}{2} = \frac{-64 \pm 8\sqrt{73}}{2} = -32 \pm 4\sqrt{73}\n]", "Recall $ u = a^2 $, and since $ a^2 \geq 0 $, only the positive root is valid:", "[\na^2 = -32 + 4\sqrt{73}\n]", "We now check whether $ -32 + 4\sqrt{73} > 0 $:\nSince $ \sqrt{73} \approx 8.544 $, then $ 4\sqrt{73} \approx 34.176 $, so $ -32 + 34.176 = 2.176 > 0 $. Valid.", "---", "### Solve for $ a $", "Take square roots:", "[\na = \pm \sqrt{-32 + 4\sqrt{73}} = \pm \sqrt{4(\sqrt{73} - 8)} = \pm 2\sqrt{\sqrt{73} - 8}\n]", "---", "### Final Result", "The solutions to the equation $ a^2 - \frac{144}{a^2} = -64 $ are:", "[\na = \pm 2\sqrt{\sqrt{73} - 8}\n]", "---", "### Why This Strategy Works", "This problem demonstrates how clever substitution transforms a complex rational equation into a quadratic, dramatically simplifying solution efforts. By replacing $ a^2 $ with $ u $, we reduce an equation involving inverses and powers into a clean quadratic form—ideal for algebraic manipulation and numerical approximation when necessary.", "Whether in algebra, calculus, or applied mathematics, recognizing substitution opportunities is key to efficient problem-solving.", "---", "### Key Takeaways", "- Multiplying through by $ a^2 $ removes denominators cleanly.\n- Rearranging allows variable substitution to convert polynomials.\n- Always verify the domain (e.g., $ a^2 > 0 $) to ensure valid solutions.\n- Quadratic formulas and algebraic simplifications turn barriers into breakthroughs.", "---", "Keywords: solve $ a = 12/b $, $ a^2 - 144/a^2 = -64 $, multivariate substitution, polynomial transformation, algebra solutions, $ u = a^2 $, quadratic equation, square roots, equation solving", "Meta Description:\nLearn how to solve $ a^2 - \frac{144}{a^2} = -64 $ by multiplying by $ a^2 $, substituting $ u = a^2 $, and applying the quadratic formula. Step-by-step algebra with domain considerations and domain validation."]