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Solve for \( a \) : \[ 18 a^{2}+56 a+10=3 a^{2}+3 a \] INSTRUCTION: You may type your answers in any order. Answer: \( a=\square I \) \( a=\square \)

Ask by Watson Harper. in South Africa
Mar 08,2025

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\( a = -\frac{10}{3} \) or \( a = -\frac{1}{5} \)

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To solve for \( a \), let's start by rearranging the equation. We move all terms to one side: \[ 18a^2 + 56a + 10 - 3a^2 - 3a = 0 \] This simplifies to: \[ 15a^2 + 53a + 10 = 0 \] Next, we can use the quadratic formula: \[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 15 \), \( b = 53 \), and \( c = 10 \). First, we calculate the discriminant: \[ b^2 - 4ac = 53^2 - 4 \cdot 15 \cdot 10 = 2809 - 600 = 2209 \] Now we substitute the values into the formula: \[ a = \frac{-53 \pm \sqrt{2209}}{2 \cdot 15} \] Since \( \sqrt{2209} = 47 \): \[ a = \frac{-53 \pm 47}{30} \] Calculating the two possible values: 1. \( a = \frac{-53 + 47}{30} = \frac{-6}{30} = -\frac{1}{5} \) 2. \( a = \frac{-53 - 47}{30} = \frac{-100}{30} = -\frac{10}{3} \) So, the solutions are: \( a = -\frac{1}{5} \) \( a = -\frac{10}{3} \)

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