Question

Solve the following equation: bb−5−2b+5=50b2−25 and show your work.(4 points)

Answer 1

To solve the equation \( bb - 5 - 2b + 5 = 50b^2 - 25 \), we first rewrite \( bb \) as \( b^2 \). Thus, the equation becomes:

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

Now, simplify the left side:

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

This simplifies to:

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

Next, we move all terms to one side of the equation. Subtract \( 50b^2 - 25 \) from both sides:

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

Combining like terms gives:

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

To make it standard, multiply the entire equation by -1:

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

Now we can solve this quadratic equation using the quadratic formula:

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

Where \( A = 1 \), \( B = 2 \), and \( C = -25 \). Calculating the discriminant:

\[
B^2 - 4AC = 2^2 - 4 \cdot 1 \cdot (-25) = 4 + 100 = 104
\]

Now substitute back into the formula:

\[
b = \frac{-2 \pm \sqrt{104}}{2 \cdot 1}
\]

\(\sqrt{104}\) can be simplified to \( 2\sqrt{26} \), thus we have:

\[
b = \frac{-2 \pm 2\sqrt{26}}{2}
\]
\[
b = -1 \pm \sqrt{26}
\]

Hence, the two possible solutions for \( b \) are:

\[
b = -1 + \sqrt{26} \quad \text{and} \quad b = -1 - \sqrt{26}
\]

Therefore, the final solutions are:

\[
\boxed{-1 + \sqrt{26}} \quad \text{and} \quad \boxed{-1 - \sqrt{26}}
\]