Question

Suppose the quadratic function
`f(x)=ax^(2) +bx+c` with
`f(-2)=0` and
`-(b)/(2a) =1`. Solve
`f(x)=0`.

Answer 1

Explanation:

hello :

let : X1 and X2 this solution you have : X1 + X2 = -b/a

but given : -b/2a = 1 so : -b/a = 22

now : X1 + X2 = 2

other solution is : -2 because f(-2) =0

so : -2+X2 = 2 X2 = 4

f(x) = (x-4)(x+2) = x²-2x-8 a=1 and b=- 2 and c= -8

verify : f(-2) = (-2)²-2(-2)-8=8-8=0....right

-b/2a = -(-2)/2(1) = 1 ...right

the soultions are : 4 and -2