Find the value of k. If there is one solution. The square root of k plus x equals 52 plus x

To find the value of \( k \) such that there is exactly one solution to the equation involving square roots, let's analyze the given equation:

\[
\sqrt{k + x} = 52 + x
\]

### Step-by-Step Solution:

1. **Isolate the square root term**: The equation is already isolated in terms of the square root on the left side.

2. **Square both sides**: To eliminate the square root, square both sides of the equation.

\[
(\sqrt{k + x})^2 = (52 + x)^2
\]

This simplifies to:

\[
k + x = (52 + x)^2
\]

3. **Expand and rearrange the equation**: Expand the square on the right-hand side.

\[
k + x = 52^2 + 2 \cdot 52 \cdot x + x^2
\]

Calculating \( 52^2 \):

\[
52^2 = 2704
\]

Substitute back:

\[
k + x = 2704 + 104x + x^2
\]

Rearranging all terms to one side to form a quadratic equation:

\[
x^2 + 104x + 2704 - x - k = 0
\]

Simplify further:

\[
x^2 + 103x + (2704 - k) = 0
\]

4. **Condition for exactly one solution**: For the quadratic equation \( ax^2 + bx + c = 0 \) to have exactly one solution, its discriminant must be zero.

The discriminant \( \Delta \) for \( ax^2 + bx + c = 0 \) is given by:

\[
\Delta = b^2 - 4ac
\]

In our equation, \( a = 1 \), \( b = 103 \), and \( c = 2704 - k \). The discriminant becomes:

\[
\Delta = 103^2 - 4 \cdot 1 \cdot (2704 - k)
\]

Calculate \( 103^2 \):

\[
103^2 = 10609
\]

Thus, the discriminant is:

\[
\Delta = 10609 - 4(2704 - k)
\]

Set the discriminant to zero for exactly one solution:

\[
10609 - 4(2704 - k) = 0
\]

5. **Solve for \( k \)**:

\[
10609 = 4(2704 - k)
\]

Divide both sides by 4:

\[
2652.25 = 2704 - k
\]

Rearranging for \( k \):

\[
k = 2704 - 2652.25
\]

\[
k = 51.75
\]

### Conclusion:
The value of \( k \) such that there is exactly one solution to the equation \(\sqrt{k + x} = 52 + x\) is \(\boxed{51.75}\).

Would you like more details or have any questions?

Here are 5 related questions:
1. What is the nature of the solutions if \( k \) is not equal to 51.75?
2. How does the value of \( k \) affect the graph of \(\sqrt{k + x}\) compared to the line \(52 + x\)?
3. Can there be no solutions to the equation? If yes, under what condition?
4. What would the solution look like if we had \(\sqrt{k + x} = ax + b\) instead?
5. How do you determine the domain of the function \(\sqrt{k + x}\)?

**Tip:** Always remember that squaring both sides of an equation may introduce extraneous solutions. It is important to check potential solutions in the original equation.

Discover how to find the value of k in the equation √(k + x) = 52 + x with a single solution. This problem involves solving a quadratic equation, understanding the discriminant, and applying algebraic manipulation. Suitable for advanced high school students.

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