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State the possible values of k that would make the following expression a perfect square trinomial kx^2 - 198x + 81.

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Final answer:

The value of k that makes the expression kx^2 - 198x + 81 a perfect square trinomial is 11, provided by setting 2ab = -198 and b^2 = 81, solving for k and b simultaneously.

Step-by-step explanation:

To determine the possible values of k that would make the expression kx^2 - 198x + 81 a perfect square trinomial, we need to find a value for k so that the x-term can be expressed as the square of half the coefficient of the x-term.

A perfect square trinomial takes the form (ax+b)^2 = a^2x^2 + 2abx + b^2. To make the given expression a perfect square trinomial, 2ab must equal -198 and b^2 must equal 81. By taking the square root of 81, we get b = ±9. Since we need a negative middle term (because our middle term is -198), we'll use -9 for b.

k would then be computed using the relationship 2ab = -198, giving us k = -198/(2*(-9)) which simplifies to k = -198/-18 = 11. As such, k must be 11 for the expression to be a perfect square trinomial.

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