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Solve the equation. \[ \frac{5}{x^{2}}-\frac{54}{x}=11 \]

Ask by Patel Ramos. in the United States
Feb 03,2025

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Tutor-Verified Answer

Answer

The solutions are \( x = \frac{1}{11} \) and \( x = -5 \).

Solution

To solve the equation: \[ \frac{5}{x^{2}} - \frac{54}{x} = 11 \] **Step 1: Rewrite the Equation** First, bring all terms to one side to set the equation to zero: \[ \frac{5}{x^{2}} - \frac{54}{x} - 11 = 0 \] **Step 2: Substitute** Let \( y = \frac{1}{x} \). Then: \[ \frac{5}{x^{2}} = 5y^2 \quad \text{and} \quad \frac{54}{x} = 54y \] Substituting these into the equation: \[ 5y^2 - 54y - 11 = 0 \] **Step 3: Solve the Quadratic Equation** Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 5 \), \( b = -54 \), and \( c = -11 \): \[ y = \frac{54 \pm \sqrt{(-54)^2 - 4 \cdot 5 \cdot (-11)}}{2 \cdot 5} \] \[ y = \frac{54 \pm \sqrt{2916 + 220}}{10} \] \[ y = \frac{54 \pm \sqrt{3136}}{10} \] \[ y = \frac{54 \pm 56}{10} \] This yields two solutions for \( y \): \[ y = \frac{54 + 56}{10} = \frac{110}{10} = 11 \] \[ y = \frac{54 - 56}{10} = \frac{-2}{10} = -\frac{1}{5} \] **Step 4: Find \( x \) from \( y \)** Recall that \( y = \frac{1}{x} \), so: 1. If \( y = 11 \): \[ x = \frac{1}{11} \] 2. If \( y = -\frac{1}{5} \): \[ x = -5 \] **Step 5: Verify Solutions** Ensure that both solutions do not make the original equation undefined (i.e., \( x \neq 0 \)). Both \( x = \frac{1}{11} \) and \( x = -5 \) satisfy this condition. **Final Answer:** The solutions are \( x = \dfrac{1}{11} \) and \( x = -5 \).

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Bonus Knowledge

To solve the equation \[ \frac{5}{x^{2}} - \frac{54}{x} = 11, \] first, let's eliminate the fractions by multiplying both sides by \(x^2\) (assuming \(x \neq 0\)): \[ 5 - 54x = 11x^2. \] Rearranging the equation gives us: \[ 11x^2 + 54x - 5 = 0. \] Now, we can solve this quadratic equation using the quadratic formula, where \(a = 11\), \(b = 54\), and \(c = -5\): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \] Calculating the discriminant: \[ b^2 - 4ac = 54^2 - 4 \cdot 11 \cdot (-5) = 2916 + 220 = 3136. \] Now, taking the square root: \[ \sqrt{3136} = 56. \] Now we can substitute back into the quadratic formula: \[ x = \frac{-54 \pm 56}{2 \cdot 11}. \] Calculating the two possible values for \(x\): 1. \(x = \frac{-54 + 56}{22} = \frac{2}{22} = \frac{1}{11}\). 2. \(x = \frac{-54 - 56}{22} = \frac{-110}{22} = -5\). Thus, the solutions to the equation are: \[ x = \frac{1}{11} \quad \text{and} \quad x = -5. \] In conclusion, the solutions to the original equation are \(x = \frac{1}{11}\) and \(x = -5\).

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