To solve the radical equation
\[ \sqrt{b + 14} = \frac{4}{\sqrt{5b - 14}}, \]
we will first eliminate the square root by squaring both sides. Squaring both sides gives us:
\[ \left(\sqrt{b + 14}\right)^2 = \left(\frac{4}{\sqrt{5b - 14}}\right)^2. \]
This simplifies to:
\[ b + 14 = \frac{16}{5b - 14}. \]
Next, we will eliminate the fraction by multiplying both sides by \(5b - 14\):
\[ (b + 14)(5b - 14) = 16. \]
Now we can expand the left-hand side:
\[ 5b^2 - 14b + 70b - 196 = 16. \]
Simplifying this, we have:
\[ 5b^2 + 56b - 196 = 16. \]
Next, we move 16 to the left side:
\[ 5b^2 + 56b - 212 = 0. \]
Now, we can apply the quadratic formula \(b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\) where \(A = 5\), \(B = 56\), and \(C = -212\):
- Calculate the discriminant:
\[ B^2 - 4AC = 56^2 - 4 \cdot 5 \cdot (-212) = 3136 + 4240 = 7376. \]
- Calculate \(b\):
\[ b = \frac{-56 \pm \sqrt{7376}}{2 \cdot 5}. \]
Calculating \(\sqrt{7376}\):
\[ \sqrt{7376} = 86. \]
Now substituting back:
\[ b = \frac{-56 \pm 86}{10}. \]
This results in two potential solutions:
- \(b = \frac{30}{10} = 3\).
- \(b = \frac{-142}{10} = -14.2\).
Now let's check for extraneous solutions by substituting these back into the original equation:
- For \(b = 3\):
\[ \sqrt{3 + 14} = \sqrt{17}, \quad \frac{4}{\sqrt{5(3) - 14}} = \frac{4}{\sqrt{15 - 14}} = 4. \] Thus, \(\sqrt{17} \neq 4\) — not valid.
- For \(b = -14.2\):
\[ \sqrt{-14.2 + 14} = \sqrt{-0.2}, \] which is not valid since we cannot take the square root of a negative number in the real number system.
Therefore, neither solution is valid. The original equation has no real number solutions.