To determine the nature of the solutions for the equation
\(\frac{3}{4z} - \frac{1}{4z} + 3 = \frac{2}{4z + 5}\),
let's simplify it step by step.
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First, we can combine like terms on the left side: \[ \frac{3}{4z} - \frac{1}{4z} = \frac{3 - 1}{4z} = \frac{2}{4z}. \] So the equation now reads: \[ \frac{2}{4z} + 3 = \frac{2}{4z + 5}. \]
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Next, let's multiply every term by \(4z(4z + 5)\) to eliminate the fractions. This gives us: \[ 2(4z + 5) + 3(4z)(4z + 5) = 2(4z). \] Upon expanding, we have: \[ 8z + 10 + 12z^2 + 15z = 2(4z). \] Simplifying further: \[ 12z^2 + (8z + 15z - 8z) + 10 = 0, \] which simplifies to: \[ 12z^2 + 15z + 10 = 0. \]
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Now, we analyze the quadratic equation: \[ 12z^2 + 15z + 10 = 0. \]
To determine whether this quadratic equation has solutions, we can compute its discriminant, \(D\), given by: \[ D = b^2 - 4ac, \] where \(a = 12\), \(b = 15\), and \(c = 10\).
Calculating \(D\): \[ D = 15^2 - 4(12)(10) = 225 - 480 = -255. \]
Since the discriminant \(D\) is negative (\(-255 < 0\)), this implies that the quadratic equation has no real solutions.
Thus, the correct statement about the equation is:
It has no solution.
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