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 \( b^2 \). So, the equation becomes:

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

Next, we can simplify the left-hand side:

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

Now, we bring all terms to one side of the equation:

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

Combining like terms, we have:

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

To make the coefficients more manageable, we can multiply through by -1:

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

Now, we will use the quadratic formula:

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

In our case, \( A = 49 \), \( B = 2 \), and \( C = -25 \). First, we calculate the discriminant \( B^2 - 4AC \):

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

Now calculate the discriminant:

\[
B^2 - 4AC = 4 + 4900 = 4904
\]

Now we can use the quadratic formula to find \( b \):

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

Calculating \( \sqrt{4904} \):

\[
\sqrt{4904} = \sqrt{4 \cdot 1226} = 2\sqrt{1226}
\]

Now substituting back into the quadratic formula:

\[
b = \frac{-2 \pm 2\sqrt{1226}}{98} = \frac{-1 \pm \sqrt{1226}}{49}
\]

At this point, we can approximate \( \sqrt{1226} \):

\[
\sqrt{1226} \approx 35
\]

Thus, we have:

\[
b \approx \frac{-1 \pm 35}{49}
\]

Calculating the two possible values for \( b \):

1. \( b = \frac{-1 + 35}{49} = \frac{34}{49} \)
2. \( b = \frac{-1 - 35}{49} = \frac{-36}{49} \)

Therefore, the solutions for \( b \) are:

\[
b \approx \frac{34}{49} \quad \text{and} \quad b \approx \frac{-36}{49}
\]

So the final solutions are:

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