Question

Solve the following equation: bb−5−2b+5=50b2−25 and show your work.
no bullet points and no fracs

Answer 1

To solve the equation \( b^2 - 5 - 2b + 5 = 50b^2 - 25 \), we start by simplifying both sides.

The left-hand side can be simplified as follows:

1. Combine like terms:
\( b^2 - 2b - 5 + 5 = b^2 - 2b \).

So, the equation becomes:
\[ b^2 - 2b = 50b^2 - 25. \]

Next, we move all terms to one side of the equation to set it to zero:

2. Subtract \( 50b^2 \) and add \( 25 \) to both sides:
\[ b^2 - 2b - 50b^2 + 25 = 0. \]

Now combine like terms:
\[ -49b^2 - 2b + 25 = 0. \]

To make calculations easier, we can multiply the entire equation by -1 to change the signs:
\[ 49b^2 + 2b - 25 = 0. \]

Next, we can use the quadratic formula to find the roots of this equation. The quadratic formula is:
\[ b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}, \]
where \( A = 49 \), \( B = 2 \), and \( C = -25 \).

3. Calculate the discriminant:
\[ B^2 - 4AC = 2^2 - 4(49)(-25) = 4 + 4900 = 4904. \]

Now we find \( b \) using the quadratic formula:

4. Substitute into the formula:
\[ b = \frac{-2 \pm \sqrt{4904}}{2(49)}. \]

5. Simplify \( \sqrt{4904} \):
Since \( 4904 = 4 \cdot 1226 \), we have \( \sqrt{4904} = 2\sqrt{1226} \).

6. Substitute back:
\[ b = \frac{-2 \pm 2\sqrt{1226}}{98} = \frac{-1 \pm \sqrt{1226}}{49}. \]

Thus, the solutions for \( b \) are:
\[ b = \frac{-1 + \sqrt{1226}}{49} \quad \text{and} \quad b = \frac{-1 - \sqrt{1226}}{49}. \]

These are the values of \( b \) that satisfy the original equation.