How do you use completing the square to make a perfect square trinomial from #x^2 - 7/2x# + _?

1 Answer
Nov 14, 2015

#49/16#

Explanation:

In order to "complete the square," it's necessary to write a trinomial that is the equivalent to a binomial squared. Examples of perfect binomial squares would include #x^2-6x+9#, which is equivalent to #(x-3)^2#, and #36x^2-12x+1#, which is equivalent to #(6x-1)^2#.

Consider that #(x+a)^2=x^2+2ax+a^2#.
If, for example, we wanted to complete the square with #x^2+10x+?# we know that #10x#, the middle term, must be equal to #2ax#.
Therefore, if #10x=2ax#, we can divide by #2x# on both sides to find that #5=a#.
If we plug this back into the original form of #x^2+2ax+a^2#, we can complete the square by adding #a^2# to the end, which is #25#.

In your case, we can do the same thing. Your middle term, which must be equal to #2ax#, is #-7/(2)x#.

Thus, if #2ax=-7/(2)x#, if we divide both sides by #2x#, we see that #a=-7/4#.

Since #a=-7/4#, when we plug everything back into the form of #x^2+2ax+a^2#, we get #x^2-7/(2)x+color(red)(49/16)#.