To solve the equation \( b^2 - 5 - 2b + 5 = 50b^2 - 25 \), we can simplify and rearrange the terms.
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Combine like terms on the left side:
\[ b^2 - 2b - 5 + 5 = b^2 - 2b \]
So the equation simplifies to:
\[ b^2 - 2b = 50b^2 - 25 \]
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Now, bring all terms to one side of the equation:
\[ b^2 - 2b - 50b^2 + 25 = 0 \]
Combine the \( b^2 \) terms:
\[ -49b^2 - 2b + 25 = 0 \]
Multiplying the entire equation by -1 to make coefficients positive:
\[ 49b^2 + 2b - 25 = 0 \]
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Now, we can use the quadratic formula to solve for \( b \):
The quadratic formula is:
\[ b = \frac{{-B \pm \sqrt{{B^2 - 4AC}}}}{2A} \]
For our equation \( 49b^2 + 2b - 25 = 0 \):
- \( A = 49 \)
- \( B = 2 \)
- \( C = -25 \)
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Calculate \( B^2 - 4AC \):
\[ B^2 - 4AC = 2^2 - 4 \cdot 49 \cdot (-25) = 4 + 4900 = 4904 \]
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Now substitute into the quadratic formula:
\[ b = \frac{{-2 \pm \sqrt{4904}}}{2 \cdot 49} \]
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Simplifying further, the square root of 4904:
\[ \sqrt{4904} = 2\sqrt{1226} \]
Therefore, the equation becomes:
\[ b = \frac{{-2 \pm 2\sqrt{1226}}}{98} \]
Simplifying this gives:
\[ b = \frac{{-1 \pm \sqrt{1226}}}{49} \]
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Finding the approximate values for \( b \):
First, calculate \( \sqrt{1226} \approx 34.95 \).
So we have two possible solutions:
\[ b_1 \approx \frac{{-1 + 34.95}}{49} = \frac{33.95}{49} \approx 0.692 \]
\[ b_2 \approx \frac{{-1 - 34.95}}{49} = \frac{-35.95}{49} \approx -0.734 \]
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The approximate solutions for \( b \) are:
\[ b \approx 0.692 \quad \text{or} \quad b \approx -0.734 \]
Thus, the solutions to the equation \( b^2 - 5 - 2b + 5 = 50b^2 - 25 \) are approximately \( b \approx 0.692 \) and \( b \approx -0.734 \).