\Rightarrow rac{1 + r + r^2 + r^3}{r^{1.5}} = 6

["Solving \frac{1 + r + r^2 + r^3}{r^{1.5}} = 6: A Step-by-Step Guide & Key Insights", "Mathematics often presents us with elegant equations that challenge our algebraic intuition. One such equation is:", "[\n\dfrac{1 + r + r^2 + r^3}{r^{1.5}} = 6\n]", "This equation appears in various scientific, financial, and engineering contexts where power-law relationships model growth, decay, or scaling phenomena. In this article, we explore how to solve this equation step by step and uncover the meaningful values of ( r ) that satisfy it.", "---", "### Understanding the Equation", "We begin by analyzing the left-hand side:", "[\n\dfrac{1 + r + r^2 + r^3}{r^{1.5}} = 6\n]", "This expression combines a polynomial numerator ( 1 + r + r^2 + r^3 ) in the numerator and a fractional power ( r^{1.5} ) in the denominator. The equation asks: for what value(s) of ( r > 0 ) does this ratio equal 6?", "Note: Since ( r^{1.5} = r^{3/2} ), the equation is defined only for ( r > 0 ).", "---", "### Step 1: Simplify the Numerator", "The numerator is a geometric series:", "[\n1 + r + r^2 + r^3 = \dfrac{r^4 - 1}{r - 1} \quad \ ext{for } r <br/>\ne 1\n]", "However, substituting the closed form directly may complicate solving, so we proceed differently.", "First, factor numerator and rewrite the expression:", "[\n\dfrac{1 + r + r^2 + r^3}{r^{1.5}} = \dfrac{r^3 + r^2 + r + 1}{r^{3/2}} = r^{1.5} + r^{0.5} + r^{-0.5} + r^{-1.5}\n]", "This simplification transforms the ratio into a symmetric function in powers of ( r^{0.5} ).", "Let’s make a substitution to make the equation solvable.", "---", "### Step 2: Substitution to Simplify", "Let:", "[\nx = r^{0.5} \quad \Rightarrow \quad r = x^2 \quad \Rightarrow \quad r^{0.5} = x, \quad r^{1.5} = x^3, \quad r^{-0.5} = \dfrac{1}{x}, \quad r^{-1.5} = \dfrac{1}{x^3}\n]", "Then:", "[\nr^{1.5} + r^{0.5} + r^{-0.5} + r^{-1.5} = x^3 + x + \dfrac{1}{x} + \dfrac{1}{x^3}\n]", "So the equation becomes:", "[\nx^3 + x + \dfrac{1}{x} + \dfrac{1}{x^3} = 6\n]", "Multiply both sides by ( x^3 ) (since ( x > 0 ), this is valid):", "[\nx^6 + x^4 + x^2 + 1 = 6x^3\n]", "---", "### Step 3: Rearranging into a Polynomial Equation", "Bring all terms to one side:", "[\nx^6 + x^4 - 6x^3 + x^2 + 1 = 0\n]", "We now must solve this degree-6 polynomial. Look for rational roots using Rational Root Theorem. Try small integers ( x = 1 ):", "[\n1 + 1 - 6 + 1 + 1 = -2 <br/>\ne 0\n]", "Try ( x = 2 ):", "[\n64 + 16 - 48 + 4 + 1 = 37 <br/>\ne 0\n]", "Try ( x = 1.5 ):", "[\n1.5^6 = 11.390625,\quad 1.5^4 = 5.0625,\quad 6 \cdot 1.5^3 = 6 \cdot 3.375 = 20.25\n]\n[\n11.39 + 5.06 - 20.25 + 2.25 + 1 = -0.65 \quad (\ ext{negative})\n]", "Try ( x = 1.6 ):", "[\n1.6^2 = 2.56,\quad 1.6^3 = 4.096,\quad 1.6^4 = 6.5536,\quad 1.6^6 = 16.777216\n]\n[\n16.78 + 6.55 - 24.58 + 2.56 + 1 = 2.31 \quad (\ ext{positive})\n]", "So between ( x = 1.5 ) and ( x = 1.6 ), the polynomial crosses zero. But rather than approximate, let's reconsider the symmetry.", "---", "### Step 4: Exploit Symmetry in the Original Expression", "Recall:", "[\nx^3 + x + \frac{1}{x} + \frac{1}{x^3}\n= \left( x^3 + \frac{1}{x^3} \right) + \left( x + \frac{1}{x} \right)\n]", "Use identity:", "[\nx^3 + \frac{1}{x^3} = \left( x + \frac{1}{x} \right)^3 - 3\left( x + \frac{1}{x} \right)\n]", "Let ( y = x + \dfrac{1}{x} ). Since ( x > 0 ), ( y \ge 2 ) by AM-GM.", "Then:", "[\nx^3 + \frac{1}{x^3} = y^3 - 3y\n]", "So the full expression becomes:", "[\n(y^3 - 3y) + y = y^3 - 2y\n]", "Set equal to 6:", "[\ny^3 - 2y = 6\n]\n[\ny^3 - 2y - 6 = 0\n]", "Now solve for ( y ):", "Try ( y = 2 ): ( 8 - 4 - 6 = -2 )\nTry ( y = 3 ): ( 27 - 6 - 6 = 15 )\nTry ( y = 2.5 ): ( 15.625 - 5 - 6 = 4.625 )\nTry ( y = 2.2 ): ( 10.648 - 4.4 - 6 = 0.248 )\nTry ( y = 2.19 ): ( 2.19^3 \approx 10.511, \quad 10.511 - 4.38 - 6 = 0.131 )\nTry ( y = 2.18 ): ( 2.18^3 \approx 10.368, \quad 10.368 - 4.36 - 6 = 0.008 )", "Close to ( y \approx 2.18 )", "Use rational approximation: try ( y = 2 ) too low, we need better.", "Alternatively, test if ( y = \sqrt{3} )? Not helpful.", "But notice: suppose ( y = \sqrt{6} )? No.", "Actually, try rational root of ( y^3 - 2y - 6 = 0 ). Possible rational roots: ±1, ±2, ±3, ±6 — none work. So irrational.", "Use numerical methods or factor.", "But observe: try ( y = \sqrt[3]{6 + 2y} ) — not helpful.", "Alternatively, use substitution back:", "We suspect one real solution ( y > 2 ). Set ( f(y) = y^3 - 2y - 6 ). Since ( f(2) = -2 ), ( f(3) = 15 ), and ( f ) increasing, one real root ( y \approx 2.18 )", "Use Newton-Raphson:", "( f(y) = y^3 - 2y - 6 )\n( f'(y) = 3y^2 - 2 )", "Start with ( y_0 = 2.18 ):", "( f(2.18) = 2.18^3 = 10.368, -\ 4.36, -6 → 10.368 - 4.36 - 6 = 0.008 ) — very close", "Try ( y = 2.175 ):", "( 2.175^2 = 4.730625 ),\n( 2.175^3 = 2.175 \cdot 4.730625 \approx 10.285 )\nThen ( 10.285 - 2(2.175) - 6 = 10.285 - 4.35 - 6 = -0.065 )", "So root between 2.175 and 2.18", "Interpolate: zero at ( y \approx 2.178 )", "Now recall ( y = x + \frac{1}{x} ), ( x > 0 )", "For ( x + \frac{1}{x} = y ), solve quadratic:", "[\nx^2 - yx + 1 = 0 \Rightarrow x = \dfrac{y \pm \sqrt{y^2 - 4}}{2}\n]", "Since ( y \approx 2.18 > 2 ), real solutions exist.", "Take positive root:", "[\nx = \dfrac{2.18 + \sqrt{(2.18)^2 - 4}}{2} \approx \dfrac{2.18 + \sqrt{4.75 - 4}}{2} = \dfrac{2.18 + \sqrt{0.75}}{2} \approx \dfrac{2.18 + 0.866}{2} \approx \dfrac{3.046}{2} \approx 1.523\n]", "Then ( r = x^2 \approx (1.523)^2 \approx 2.32 )", "But better: use exact expression.", "---", "### Step 5: Exact Solution Approach", "We return to:", "[\ny^3 - 2y - 6 = 0\n]", "Let’s use trigonometric substitution for depressed cubic.", "For ( y^3 + py + q = 0 ), here ( p = -2 ), ( q = -6 )", "Discriminant: ( D = \left( \dfrac{q}{2} \right)^2 + \left( \dfrac{p}{3} \right)^3 = (-3)^2 + \left( -\dfrac{2}{3} \right)^3 = 9 - \dfrac{8}{27} = \dfrac{243 - 8}{27} = \dfrac{235}{27} > 0 )", "So one real root:", "[\ny = \sqrt[3]{-\dfrac{q}{2} + \sqrt{ \left( \dfrac{q}{2} \right)^2 + \left( \dfrac{p}{3} \right)^3 }} + \sqrt[3]{-\dfrac{q}{2} - \sqrt{ \left( \dfrac{q}{2} \right)^2 + \left( \dfrac{p}{3} \right)^3 }}\n]", "[\n= \sqrt[3]{3 + \sqrt{ \dfrac{235}{27} }} + \sqrt[3]{3 - \sqrt{ \dfrac{235}{27} }}\n]", "This is exact but messy. Instead, accept numerical solution.", "But notice: we can guess rational solution? Try ( y = 3 ): too big. ( y = \sqrt{6} \approx 2.45 )? ( 2.45^3 \approx 14.6 ), too big.", "Wait — go back.", "Try to factor original polynomial:", "[\nx^6 + x^4 - 6x^3 + x^2 + 1\n]", "Try factoring as quadratic in ( x^3 ): not helpful.", "Try factoring into two cubics:", "Assume:", "[\n(x^3 + a x^2 + b x + c)(x^3 - a x^2 + d x + e) = x^6 + x^4 - 6x^3 + x^2 + 1\n]", "But too long.", "Instead, go back to the expression:", "[\nx^3 + x + r^{-0.5} + r^{-1.5} = 6\n]", "But earlier transformation gave us a solvable cubic in ( y = x + x^{-1} ):", "[\ny^3 - 2y = 6\n]", "Now solve numerically with precision.", "Let ( f(y) = y^3 - 2y - 6 )", "Use Newton-Raphson:\n( f'(y) = 3y^2 - 2 )\nStart ( y_0 = 2.18 )\n( f(2.18) = 2.18^3 = 10.368072 - 4.36 - 6 = 0.008072 )\n( f'(2.18) = 3(4.7524) - 2 = 14.2572 - 2 = 12.2572 )\n( y_1 = 2.18 - \dfrac{0.008072}{12.2572} \approx 2.18 - 0.000659 \approx 2.1793 )", "Compute:\n( y = 2.1793 ), ( y^3 \approx 2.1793^3 \approx 10.334 )\n( 2y = 4.3586 ), ( f(y) = 10.334 - 4.3586 - 6 = -0.0246 ) — too low?", "Better:\n( 2.18^3 = 10.368072 )\n( 2y = 4.36 )\n( f = 10.368 - 4.36 - 6 = 0.008 ) — good", "Derivative: ( 3(2.18)^2 = 34.7524 = 14.2572 )\nError: 0.008 → correction: ( \Delta y = -0.008 / 14.2572 \approx 0.0005616 )\nSo ( y \approx 2.18 - 0.00056 = 2.17944 )", "Now compute ( r = x^2 ), where ( x + 1/x = y \approx 2.17944 )", "Solve: ( x^2 - 2.17944 x + 1 = 0 )\nDiscriminant: ( (2.17944)^2 - 4 \approx 4.7518 - 4 = 0.7518 )\n( x = \dfrac{2.17944 \pm \sqrt{0.7518}}{2} \approx \dfrac{2.17944 \pm 0.8662}{2} )", "Take larger root (since ( r > 1 ) likely):\n( x \approx \dfrac{2.17944 + 0.8662}{2} = \dfrac{3.04564}{2} = 1.52282 )", "Then ( r = x^2 \approx (1.52282)^2 \approx 2.319 )", "Check in original:", "Let ( r \approx 2.319 ), ( r^{1.5} = r \sqrt{r} \approx 2.319 \ imes \sqrt"]