Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-2) = 0, then b + c =
Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-2) = 0, then b + c =
Answer/Solution
-16
Steps/Work
f(x) = x^2 + bx + c. If f(6) = 0 and f(-2) = 0, then b + c =
f(6) = 0= 36+6b+c--- Taking 36 to the other side
-> 6b+c= -36
f(-2) =0= 4-2b+c --- taking -2b+c to the other side
-> 2b-c=4
When we add these 2 equations, we get 8b= -32---> b= -4
And while substituting b= -4 we get c= -12.
b+c= -16--- Answer C
f(6) = 0= 36+6b+c--- Taking 36 to the other side
-> 6b+c= -36
f(-2) =0= 4-2b+c --- taking -2b+c to the other side
-> 2b-c=4
When we add these 2 equations, we get 8b= -32---> b= -4
And while substituting b= -4 we get c= -12.
b+c= -16--- Answer C