For what value of #k# is #x^2-21x+k# a perfect square trinomial?

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Answer 1 · 2017-03-26T08:09:21

#k = 441/4 = 110.25#

Suppose:

#x^2-21x+k = (x+e)^2#

#color(white)(x^2-21x+k) = x^2+2ex+e^2#

Equating coefficients, we have:

#2e=-21#

So:

#e = -21/2#

and

#k = e^2 = (-21/2)^2 = 441/4 = 110.25#