Find the real solutions of the following equation.
If k equals StartFraction x plus 5 Over x minus 5 EndFraction
and k squared minus k equals 30​, find x.
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Part 1
Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.
A.
The solution set is StartSet nothing EndSet

To solve the given problem, let's start by analyzing the equations provided:

1. \( k = \frac{x + 5}{x - 5} \)
2. \( k^2 - k = 30 \)

We need to find the value of \( x \) that satisfies these equations. 

### Step 1: Express \( k \) in terms of \( x \)

From the first equation, we have:

\[
k = \frac{x + 5}{x - 5}
\]

### Step 2: Substitute \( k \) into the second equation

We know:

\[
k^2 - k = 30
\]

Substitute \( k = \frac{x + 5}{x - 5} \):

\[
\left(\frac{x + 5}{x - 5}\right)^2 - \frac{x + 5}{x - 5} = 30
\]

### Step 3: Let \( k = \frac{x + 5}{x - 5} \)

\[
\frac{(x + 5)^2}{(x - 5)^2} - \frac{x + 5}{x - 5} = 30
\]

### Step 4: Common Denominator

To simplify, find a common denominator:

\[
\frac{(x + 5)^2 - (x + 5)(x - 5)}{(x - 5)^2} = 30
\]

Expanding the terms in the numerator:

\[
\frac{x^2 + 10x + 25 - (x^2 - 25)}{(x - 5)^2} = 30
\]

\[
\frac{x^2 + 10x + 25 - x^2 + 25}{(x - 5)^2} = 30
\]

\[
\frac{10x + 50}{(x - 5)^2} = 30
\]

### Step 5: Simplify the equation

\[
\frac{10(x + 5)}{(x - 5)^2} = 30
\]

\[
\frac{x + 5}{(x - 5)^2} = 3
\]

### Step 6: Cross-multiply to solve for \( x \)

\[
x + 5 = 3(x - 5)^2
\]

Expand the right side:

\[
x + 5 = 3(x^2 - 10x + 25)
\]

\[
x + 5 = 3x^2 - 30x + 75
\]

### Step 7: Bring all terms to one side to form a quadratic equation

\[
0 = 3x^2 - 31x + 70
\]

### Step 8: Solve the quadratic equation

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -31 \), and \( c = 70 \):

\[
x = \frac{-(-31) \pm \sqrt{(-31)^2 - 4 \cdot 3 \cdot 70}}{2 \cdot 3}
\]

\[
x = \frac{31 \pm \sqrt{961 - 840}}{6}
\]

\[
x = \frac{31 \pm \sqrt{121}}{6}
\]

\[
x = \frac{31 \pm 11}{6}
\]

### Step 9: Calculate the solutions

1. \( x = \frac{31 + 11}{6} = \frac{42}{6} = 7 \)
2. \( x = \frac{31 - 11}{6} = \frac{20}{6} = \frac{10}{3} \)

### Step 10: Check for restrictions

Since \( k = \frac{x+5}{x-5} \) involves division by \( x - 5 \), \( x = 5 \) is not allowed. Both solutions \( x = 7 \) and \( x = \frac{10}{3} \) do not violate this restriction.

### Conclusion

The real solutions are:

\[
\boxed{x = 7 \text{ and } x = \frac{10}{3}}
\]

Would you like further details or have any questions?

Here are some related questions you might explore:

1. How do you solve similar equations involving fractions and squares?
2. What happens if \( x = 5 \) is substituted into the original equation?
3. How do you determine restrictions when dealing with rational functions?
4. What are the steps to factor a quadratic equation effectively?
5. How can quadratic equations be used to model real-world scenarios?

**Tip:** Always check your solutions by substituting them back into the original equations to verify their validity.

Learn how to find real solutions of the rational function k = (x + 5) / (x - 5) using algebraic techniques. This problem involves solving a quadratic equation derived from the given conditions. Suitable for advanced high school students and beyond.

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