(c - 3)square = csquare + kc + 9. what is the value of k?

We use the caret (^) to indicate exponentiation.

... (c-3)^2 = c^2 + kc + 9
or c^2 - 6c + 9 = c^2 + kc + 9
or k = -6

the value of k is -6 by solving eqn.
Acc. 2 ques <c-3>square=c square+kc+9
=> c square + 9 - 6c = c square +kc + 9
=> -6c =kc
=> k= -6
then the value of k is -6
okkkkkk

(c - 3)^2 = c^2 + kc + 9
(c - 3)(c - 3) = c^2 + kc + 9
c(c) - c(3) - 3(c) - 3(-3) = c^2 + kc + 9
c^2 - 3c - 3c - (-9) = c^2 + kc + 9
c^2 - 6c + 9 = c^2 + kc + 9
k = -6

(c-3)^2 = (c-3)(c-3) = c(c-3) - 3(c-3) = c^2 - 3c - 3c + 9 = c^2 - 6c + 9

(c – 3)² = c² + kc + 9
you have two unknowns and one equation, and this normally cannot be solved. but it seems that all the c's cancel out.

kc = (c – 3)² – c² – 9
kc = c² – 6c + 9 – c² – 9
kc = – 6c
k = – 6

.

(c-3)^2 = c^2 + k*c + 9
i.e, c^2 - 6*c + 9 = c^2 + k*c + 9
therefore, k = -6

-6

(c-3)^2 = c^2 + kc +9
c^2 - 6c + 9 = c^2 + kc + 9

-6 = k