Let $ y = \frac{1}{x} $, then from $ 14x^2 + 42x + 27 = 0 $, divide by $ x^2 $:

["Understanding the Quadratic Equation Using Inverse Function: Solving $14x^2 + 42x + 27 = 0$ by Dividing by $x^2$", "When solving quadratic equations, especially those involving rational functions, a powerful technique is dividing the entire equation by $x^2$, particularly when $x <br/>\neq 0$. Let’s explore how this method simplifies solving $14x^2 + 42x + 27 = 0$ using the substitution $y = \frac{1}{x}$, a valuable approach in algebraic problem-solving.", "### Step 1: Start with the given quadratic equation", "We begin with:\n$$\n14x^2 + 42x + 27 = 0\n$$", "For values of $x <br/>\neq 0$, divide every term by $x^2$:\n$$\n\frac{14x^2}{x^2} + \frac{42x}{x^2} + \frac{27}{x^2} = 0\n$$\nSimplifying gives:\n$$\n14 + \frac{42}{x} + \frac{27}{x^2} = 0\n$$", "### Step 2: Apply the substitution $ y = \frac{1}{x} $", "Let $ y = \frac{1}{x} $, then $ \frac{1}{x} = y $ and $ \frac{1}{x^2} = y^2 $. Substitute:\n$$\n14 + 42y + 27y^2 = 0\n$$", "Rewriting in standard quadratic form:\n$$\n27y^2 + 42y + 14 = 0\n$$", "### Step 3: Solve the transformed quadratic equation", "Use the quadratic formula:\n$$\ny = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \quad \ ext{where } a = 27, b = 42, c = 14\n$$", "First, compute the discriminant:\n$$\n\Delta = 42^2 - 4 \cdot 27 \cdot 14 = 1764 - 1512 = 252\n$$", "Now compute:\n$$\ny = \frac{-42 \pm \sqrt{252}}{54}\n$$", "Simplify $ \sqrt{252} $:\n$$\n\sqrt{252} = \sqrt{36 \cdot 7} = 6\sqrt{7}\n$$", "So:\n$$\ny = \frac{-42 \pm 6\sqrt{7}}{54} = \frac{-7 \pm \sqrt{7}}{9}\n$$", "### Step 4: Recover $x$ from $y = \frac{1}{x}$", "Since $ y = \frac{1}{x} $, then $ x = \frac{1}{y} $. Substitute both solutions:", "$$\nx = \frac{1}{\frac{-7 + \sqrt{7}}{9}} = \frac{9}{-7 + \sqrt{7}}, \quad x = \frac{1}{\frac{-7 - \sqrt{7}}{9}} = \frac{9}{-7 - \sqrt{7}}\n$$", "To rationalize the denominators, multiply numerator and denominator by the conjugate:", "For $ x_1 = \frac{9}{-7 + \sqrt{7}} \cdot \frac{-7 - \sqrt{7}}{-7 - \sqrt{7}} = \frac{9(-7 - \sqrt{7})}{(-7)^2 - (\sqrt{7})^2} = \frac{9(-7 - \sqrt{7})}{49 - 7} = \frac{9(-7 - \sqrt{7})}{42} = \frac{-3(7 + \sqrt{7})}{14} $", "Similarly,\n$$\nx_2 = \frac{9(-7 + \sqrt{7})}{42} = \frac{-3(7 - \sqrt{7})}{14}\n$$", "### Summary", "Dividing the equation $14x^2 + 42x + 27 = 0$ by $x^2$ and substituting $y = \frac{1}{x}$ transforms the problem into a cleaner quadratic in $y$, enabling efficient solution via the quadratic formula. The final solutions are:\n$$\nx = \frac{-3(7 + \sqrt{7})}{14}, \quad x = \frac{-3(7 - \sqrt{7})}{14}\n$$", "### Why This Method Works", "Dividing by $x^2$ leverages symmetry in the equation when input values are inverted, turning a potentially messy quadratic into a standard form. The substitution $ y = \frac{1}{x} $ simplifies evaluation and avoids complex sign handling in original variable form — especially useful in Olympiad-level or advanced algebra contexts.", "---", "### SEO Keywords:\nsolve $14x^2 + 42x + 27 = 0$, divide quadratic by $x^2$, substitution $y = \frac{1}{x}$, rational function technique, quadratic solution method, algebraic substitution, inverse variable substitution, algebra problem-solving", "Use this approach next time you face a quadratic with linear and constant terms — it simplifies solutions elegantly."]