Question

Solve the following equation: bb−5−2b+5=50b2−25
𝑏
𝑏
−
5
−
2
𝑏
+
5
=
50
𝑏
2
−
25
and show your work.

Answer 1

To solve the equation

\[
bb - 5 - 2b + 5 = 50b^2 - 25,
\]

let's rewrite it clearly. It appears you meant \( b^2 - 2b - 5 = 50b^2 - 25 \). If that's correct, we can solve the equation as follows:

1. Start with the equation:

\[
b^2 - 2b - 5 = 50b^2 - 25.
\]

2. Move all terms to one side of the equation by subtracting \( 50b^2 \) and adding \( 25 \):

\[
b^2 - 50b^2 - 2b - 5 + 25 = 0.
\]

3. Combine like terms:

\[
-49b^2 - 2b + 20 = 0.
\]

4. Multiply the entire equation by -1 to simplify:

\[
49b^2 + 2b - 20 = 0.
\]

5. Now, we can use the quadratic formula \( b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \) where \( A = 49, B = 2, C = -20 \).

6. Calculate the discriminant \( B^2 - 4AC \):

\[
B^2 - 4AC = 2^2 - 4 \cdot 49 \cdot (-20).
\]

\[
= 4 + 3920 = 3924.
\]

7. Now, calculate \( \sqrt{3924} \):

Since \( 3924 = 4 \times 981 \) and \( \sqrt{4} = 2 \), we have:

\[
\sqrt{3924} = 2\sqrt{981}.
\]

Since \( 981 = 3^2 \times 109 \), rewrite it:

\[
\sqrt{981} = 3\sqrt{109}.
\]

So

\[
\sqrt{3924} = 2 \times 3\sqrt{109} = 6\sqrt{109}.
\]

8. Substitute back into the quadratic formula:

\[
b = \frac{-2 \pm 6\sqrt{109}}{2 \times 49} = \frac{-2 \pm 6\sqrt{109}}{98} = \frac{-1 \pm 3\sqrt{109}}{49}.
\]

9. Thus, the two solutions for \( b \) are:

\[
b = \frac{-1 + 3\sqrt{109}}{49} \quad \text{and} \quad b = \frac{-1 - 3\sqrt{109}}{49}.
\]

These are the solutions to the given equation.