Question

find the value of k that makes x^2 + (k+7)x+(7k+1) a perfect square trinomial

Answer 1

If we had
\[x^2+bx+c\]
and we wanted to find c in terms of b so that \[x^2+bx+c\] is a complete square
we would need \[c=(\frac{b}{2})^2\]
So we have that:
\[x^2+bx+c=x^2+bx+c=(x+\frac{b}{2})^2 \text{ since } c=(\frac{b}{2})^2 \]
----------------------------
So for you problem what should 7k+1 be in order for \[x^2+(k+7)x+(7k+1)\]
for this to be a complete square?

Answer 2

what is b here and what is c here based on what I have above?

Answer 3

@alexeis_nicole are you there?

Answer 4

well once you determine what b and c are just plug them into
\[c=(\frac{b}{2})^2\]
and solve for k

Answer 5

@myininaya sorry.
b = k+7
c = 7k+1

OH ! okay i get it now :)

Answer 6

Great
\[7k+1=(\frac{k+7}{2})^2\]
So you know to solve this for k right?