Question

Solve the radical equation. Check for extraneous solutions.



b
+
14
4
=
5
b
−
14
4
4

b+14
​
=
4

5b−14
​




b
=
b=

Answer 1

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 \(C = -212\):

1. Calculate the discriminant:

\[
B^2 - 4AC = 56^2 - 4 \cdot 5 \cdot (-212) = 3136 + 4240 = 7376.
\]

2. 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:

1. \(b = \frac{30}{10} = 3\).
2. \(b = \frac{-142}{10} = -14.2\).

Now let's check for extraneous solutions by substituting these back into the original equation:

1. 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**.

2. 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.