To solve the equation
\[ \frac{b}{b - 5} - \frac{2b + 5}{50b^2 - 25} = 0, \]
we can start by simplifying the terms.
First, let's simplify \(50b^2 - 25\). Note that this can be factored as:
\[ 50b^2 - 25 = 25(2b^2 - 1) = 25(2b - 1)(2b + 1). \]
Now, the equation becomes:
\[ \frac{b}{b - 5} - \frac{2b + 5}{25(2b - 1)(2b + 1)} = 0. \]
To solve this, we can first clear the fraction. To do this, we will multiply every term by the common denominator, which is \(25(b - 5)(2b - 1)(2b + 1)\):
\[ 25(b - 5)(2b - 1)(2b + 1) \times \frac{b}{b - 5} - 25(b - 5)(2b - 1)(2b + 1) \times \frac{2b + 5}{25(2b - 1)(2b + 1)} = 0. \]
This reduces to:
\[ 25b(2b - 1)(2b + 1) - (b - 5)(2b + 5) = 0. \]
We can further simplify the left-hand side:
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Calculate \(25b(2b - 1)(2b + 1)\):
\[ 25b (4b^2 - 1) = 100b^3 - 25b. \]
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Calculate \((b - 5)(2b + 5)\):
\[ (b - 5)(2b + 5) = 2b^2 + 5b - 10b - 25 = 2b^2 - 5b - 25. \]
Now substitute these back into the equation:
\[ 100b^3 - 25b - (2b^2 - 5b - 25) = 0, \]
which simplifies to:
\[ 100b^3 - 25b - 2b^2 + 5b + 25 = 0. \]
Combine like terms:
\[ 100b^3 - 2b^2 - 20b + 25 = 0. \]
Now we have a cubic equation:
\[ 100b^3 - 2b^2 - 20b + 25 = 0. \]
To find the roots of this cubic equation, we can use numerical methods or synthetic division if we suspect a rational root. Let's evaluate possible rational roots using the Rational Root Theorem.
We might try \(b = 1\):
\[ 100(1)^3 - 2(1)^2 - 20(1) + 25 = 100 - 2 - 20 + 25 = 103 \quad \text{(not a root)}. \]
Next, try \(b = -1\):
\[ 100(-1)^3 - 2(-1)^2 - 20(-1) + 25 = -100 - 2 + 20 + 25 = -57 \quad \text{(not a root)}. \]
Next, try \(b = 0.5\):
\[ 100(0.5)^3 - 2(0.5)^2 - 20(0.5) + 25 = 100(0.125) - 2(0.25) - 10 + 25 \ = 12.5 - 0.5 - 10 + 25 = 27 \quad \text{(not a root)}. \]
After attempting several potential rational roots, if we have not found one, it may be necessary to resort to numerical methods or using graphing techniques.
Assuming further analysis yields confirmed rational roots or using a numerical method shows the following roots approximations:
The original equation can be solved numerically to identify roots around values of \(b\).
Let’s summarize the approach with sufficient detail, and you would typically use a software tool if exact forms fail to identify specific values, leads to approximate values at specific boundaries where higher order polynomials are involved.
Once roots are identified, we can back substitute to ensure the values do not create undefined expressions in the original equation.
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