Question

Bowling radical 12 + 2 radical 27 = 14 radical BL

Answer 1

To solve the equation \( 12 + 2 \sqrt{27} = 14 \sqrt{BL} \), we will first simplify the left-hand side.

1. **Simplify \( \sqrt{27} \):**
\[
\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}
\]
Therefore, \( 2\sqrt{27} \) becomes:
\[
2\sqrt{27} = 2 \cdot 3\sqrt{3} = 6\sqrt{3}
\]

2. **Substituting back into the equation:**
\[
12 + 6\sqrt{3} = 14 \sqrt{BL}
\]

3. **Rearranging to isolate \( \sqrt{BL} \):**
Let’s divide all terms by 14:
\[
\frac{12}{14} + \frac{6\sqrt{3}}{14} = \sqrt{BL}
\]

4. **Simplify the fractions:**
\[
\frac{12}{14} = \frac{6}{7}, \quad \text{and} \quad \frac{6\sqrt{3}}{14} = \frac{3\sqrt{3}}{7}
\]
Thus, we have:
\[
\sqrt{BL} = \frac{6}{7} + \frac{3\sqrt{3}}{7}
\]
This can be combined:
\[
\sqrt{BL} = \frac{6 + 3\sqrt{3}}{7}
\]

5. **Squaring both sides:**
To find \( BL \), we square both sides:
\[
BL = \left( \frac{6 + 3\sqrt{3}}{7} \right)^2
\]
Expanding:
\[
= \frac{(6 + 3\sqrt{3})^2}{49}
\]
\[
(6 + 3\sqrt{3})^2 = 6^2 + 2 \cdot 6 \cdot 3\sqrt{3} + (3\sqrt{3})^2
\]
\[
= 36 + 36\sqrt{3} + 27 = 63 + 36\sqrt{3}
\]
Thus:
\[
BL = \frac{63 + 36\sqrt{3}}{49}
\]

So, the final answer is:
\[
BL = \frac{63 + 36\sqrt{3}}{49}
\]