The given simultaneous equations are:

$x + y = k \quad \text{(1)}$ $x^2 + y^2 = 18 \quad \text{(2)}$

We are tasked with finding the possible values of $k$ such that these equations have only one solution.

Step-by-step Solution:

1. Square equation (1): $(x + y)^2 = k^2$ Expanding this: $x^2 + 2xy + y^2 = k^2$ From equation (2), substitute $x^2 + y^2 = 18$: $18 + 2xy = k^2$ Rearranging this: $2xy = k^2 - 18$ Therefore: $xy = \frac{k^2 - 18}{2} \quad \text{(3)}$

2. Use the quadratic form: From equation (1), we know $x$ and $y$ are the roots of the quadratic equation: $t^2 - (x + y)t + xy = 0$ Substituting $x + y = k$ and $xy = \frac{k^2 - 18}{2}$ into the quadratic equation: $t^2 - kt + \frac{k^2 - 18}{2} = 0$

   For this quadratic equation to have only one solution, the discriminant must be zero. The discriminant $\Delta$ is given by: $\Delta = k^2 - 4 \times 1 \times \frac{k^2 - 18}{2}$ Simplifying this: $\Delta = k^2 - 2(k^2 - 18) = k^2 - 2k^2 + 36 = -k^2 + 36$ For the quadratic to have only one solution, we set $\Delta = 0$: $-k^2 + 36 = 0$ Solving for $k^2$: $k^2 = 36$ Therefore: $k = 6 \quad \text{or} \quad k = -6$

Conclusion:

The possible values of $k$ are $6$ and $-6$.

Would you like further details on any step or have additional questions?

Here are 5 related questions:

1. How does the discriminant of a quadratic equation determine the number of solutions?
2. What happens when the discriminant is positive, zero, or negative?
3. Can you generalize the method for finding the number of solutions for similar systems of equations?
4. What are other methods for solving systems of nonlinear equations?
5. How would the problem change if the second equation were $x^2 + y^2 = 25$?

Tip: Always check the discriminant when determining the number of solutions for quadratic equations.