Question

I need help on this can someone help me

Answer 1

First one;

`f(x) = {x}^(2) - 3x - 28`
As it factorises to:
f(x) = (x - 7)(x + 4)
Which, if set equal to 0, will solve to give the desired solutions like so:
f(x) = (x - 7)(x + 4) = 0
x - 7 = 0 x + 4 = 0
x = 7 x = -4

None of the other quadratics solve to give desired solutions when set equal to 0.