I will choose the simplest concise answer. work backwards to find the original problem
2006-11-07 08:56:16 · 7 answers · asked by brian80pott 1 in Science & Mathematics ➔ Mathematics
x = 4 ==> (x - 4) = 0
x = -2 ==> (x + 2) = 0
(x - 4)(x + 2) = 0
x^2 - 2x - 8 = 0
2006-11-07 09:02:57 · answer #1 · answered by DavidK93 7 · 2⤊ 0⤋
There is no unique solution for the roots you give. Inserting your roots into the quadratic eqn, ax^2 + bx +c = 0, gives
16a + 4b + c = 0 and
4a - 2b + c = 0
Two equations, three unknowns. You're out of luck.
Graphically, this is a parabola crossing the x axis at 4, -2. Any number of parabolas can cross the x axis at these points just by adjusting the curvsture and position on the grid. Your parabola is certainly one of them, but not the only one.
2006-11-07 17:23:14 · answer #2 · answered by Steve 7 · 0⤊ 0⤋
Solution : 4, -2 means that equation is (X-4)*(X+2)=0
The equation is X^2-2X-8=0.
Give me 10!
2006-11-07 17:03:18 · answer #3 · answered by George 2 · 0⤊ 0⤋
(x - 4) (x + 2) = 0
x^2 - 2x - 8 = 0
2006-11-07 17:03:51 · answer #4 · answered by bh8153 7 · 0⤊ 0⤋
x=-2 and x=4
x+2=0 and x-4=0
(x+2)(x-4)=y
After FOIL, you get y=x^2 -2x - 8
2006-11-07 17:06:27 · answer #5 · answered by peachzncream3127 2 · 0⤊ 0⤋
x^2-2x-8
2006-11-07 17:03:33 · answer #6 · answered by arbiter007 6 · 0⤊ 0⤋
(x-4)(x+2)=0
x(x-4)+2(x-4)=0
x^2-4x+2x-8=0
x^2-2x-8=0
2006-11-07 17:17:25 · answer #7 · answered by Anonymous · 0⤊ 0⤋