What will (k) be to make (x^2-18x+k) a perfect square?

Correct answer: (81)

The direct answer is option A: \(k=81\). To make \(x^2-18x+k\) a perfect square, compare it with \((x-a)^2=x^2-2ax+a^2\). The middle coefficient is \(-18\), so \(-2a=-18\), giving \(a=9\). Therefore the constant term must be \(a^2=9^2=81\). Indeed, \(x^2-18x+81=(x-9)^2\). Option A, 81, is correct. Option B, 36, is \(6^2\), but it would pair with a middle term \(-12x\), not \(-18x\). Option C, 18, is the magnitude of the middle coefficient, not its required half-square. Option D, 324, is \(18^2\), which is too large; the needed number is the square of half the coefficient, not the square of the whole coefficient. The standard rule is: for \(x^2+bx+c\) to be a square, \(c=(b/2)^2\), when the leading coefficient is 1. Exam cue: halve the middle coefficient first, keep its sign for the bracket, then square it; \((-9)^2\) is positive 81.