How do you find the c that makes the trinomial #x^2-7x+c# a perfect square?

2 Answers
Aug 19, 2017

Answer:

#c=49/4#

Explanation:

#(a - b)^2 = a^2 - 2ab + b^2#
In the above see the relation between middle term and the last term: What do you need to do to #-2ab# to become #b^2#?
divide by #2a# then square the results:
#(-2ab)/(2a) = -b# => square: #b^2#
Now in this case:
#x^2 - 7x + c#
#c = (7/2)^2 = 49/4#

Aug 19, 2017

Answer:

#c=49/4=12 1/4#

Explanation:

If a trinomial of the form #x^2+bx+c# is a perfect square, then its determinant #b^2-4c# must be equal to zero.

In #x^2-7x+c#, determinant #(-7)^2-4xxc=0# means

#49-4c=0#

i.e. #4c=49#

or #c=49/4=12 1/4#,

and then #x^2-7x+49/4=(x-7/2)^2#