Question

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

Answer 1

To solve the equation \( bb - 5 - 2b + 5 = 50b^2 - 25 \), we first rewrite it for clarity. Noting that \( bb \) can be interpreted as \( b^2 \), we can rewrite the equation as:

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

Next, let's move all terms to one side of the equation to set it equal to zero:

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

Now, combine like terms:

\[
b^2 - 50b^2 - 2b + 25 = 0
\]
\[
-49b^2 - 2b + 25 = 0
\]

To simplify solving, we can multiply the entire equation by \(-1\):

\[
49b^2 + 2b - 25 = 0
\]

Next, we can use the quadratic formula to solve for \( b \). The quadratic formula is given by:

\[
b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}
\]

In our case, \( A = 49 \), \( B = 2 \), and \( C = -25 \). We need to compute the discriminant \( B^2 - 4AC \):

\[
B^2 - 4AC = 2^2 - 4(49)(-25)
\]
\[
= 4 + 4900
\]
\[
= 4904
\]

Now substituting back into the quadratic formula:

\[
b = \frac{-2 \pm \sqrt{4904}}{2 \cdot 49}
\]

Let's simplify \( \sqrt{4904} \):
Finding the prime factorization of \( 4904 \):
- \( 4904 = 4 \times 1226 = 4 \times 2 \times 613 = 2^2 \times 2 \times 613 = 2^3 \times 613 \)

Thus, \( \sqrt{4904} = 2\sqrt{1226} \).

Continuing with the square root approximation, we can calculate:

\[
\sqrt{4904} \approx 70
\]

Now substitute this back into the formula:

\[
b = \frac{-2 \pm 70}{98}
\]

Calculating the two potential \( b \) values:

1. Using the plus sign:

\[
b = \frac{-2 + 70}{98} = \frac{68}{98} = \frac{34}{49}
\]

2. Using the minus sign:

\[
b = \frac{-2 - 70}{98} = \frac{-72}{98} = \frac{-36}{49}
\]

Thus, the solutions to the equation are:

\[
b = \frac{34}{49} \quad \text{or} \quad b = -\frac{36}{49}
\]

In summary, the solutions are:

\[
\boxed{\frac{34}{49} \text{ and } -\frac{36}{49}}
\]