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Find the real solutions to the equations. Part: \( 0 / 2 \) Part 1 of 2 (a) \( \frac{4 x^{2}-36}{x^{2}-x+4}=0 \) The solution set is

Ask by Marshall Cummings. in the United States
Mar 18,2025

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Answer

The solutions are \( x = -3 \) and \( x = 3 \).

Solution

The solution process is as follows: 1. Solve the numerator: \[ 4x^2 - 36 = 0 \quad \Longrightarrow \quad 4(x^2 - 9)= 0 \] This implies: \[ x^2 - 9 = 0 \quad \Longrightarrow \quad (x-3)(x+3)=0 \] Therefore, the candidate solutions are: \[ x=3 \quad \text{or} \quad x=-3 \] 2. Verify that these solutions do not make the denominator zero: The denominator is: \[ x^2 - x + 4 \] For \( x=3 \): \[ 3^2 - 3 + 4 = 9 - 3 + 4 = 10 \neq 0 \] For \( x=-3 \): \[ (-3)^2 - (-3) + 4 = 9 + 3 + 4 = 16 \neq 0 \] Since neither \( x=3 \) nor \( x=-3 \) makes the denominator zero, both are valid. The solution set is: \[ \{-3,\,3\} \]

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To solve the equation \( \frac{4x^{2} - 36}{x^{2} - x + 4} = 0 \), we first note that a fraction equals zero when its numerator is zero. Therefore, we set the numerator equal to zero: \[ 4x^{2} - 36 = 0 \] Solving for \(x\): \[ 4x^{2} = 36 \] \[ x^{2} = 9 \] \[ x = 3 \quad \text{or} \quad x = -3 \] Next, we need to ensure that these values do not make the denominator zero. We check the denominator: \[ x^{2} - x + 4 \] We evaluate it at \(x = 3\) and \(x = -3\): For \(x = 3\): \[ 3^{2} - 3 + 4 = 9 - 3 + 4 = 10 \quad (\text{not zero}) \] For \(x = -3\): \[ (-3)^{2} - (-3) + 4 = 9 + 3 + 4 = 16 \quad (\text{not zero}) \] Since both values do not make the denominator zero, the solution set is: \(\{3, -3\}\)

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