What is the value of #c# that makes #x^2 - 10x+c# a perfect square trinomial?

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Answer 1 · 2015-11-13T22:29:28

#25#

#(a+b)^2 = a^2+2ab+b^2#

In our case #a^2 = x^2#, so we can choose #a=x#.

#a=-x# would also be possible, but would just require a different sign for #b#.

We also have #-10x = 2ab = 2xb#.

Divide both sides by #2x# to find #b = -5#.

Then #c = b^2 = (-5)^2 = 25#

In summary:

#(x-5)^2 = x^2-10x+25#