On a Quadratic equation of the form $ax^2+bx+c$, I am given the discriminant ($\Delta=b^2-4ac$) and A.
To find B and C, I tried to solve it through $x1$ and $x2$...:
- $x1 = \frac{-b-\sqrt{\Delta}}{a}$
- $x2 = \frac{-b+\sqrt{\Delta}}{a}$
...using those properties:
- $x1 + x2 = - \frac{b}{a}$
- $x1 \times x2 = \frac{c}{a}$
Without any success...
So, is it possible to retrieve B and C?
Two additional properties:
- the discriminant is always positive
- B is always negative
One equation
$$\Delta=B^2-4AC\tag{1}$$
with 2 unknowns $B$ and $C$: there is a fundamental indeterminacy.
A counterexample among many: If $A=1/4$ and $\Delta=16$, (1) becomes:
$$16=B^2-C$$
which is possible for $(B=-4, \ C=0)$, $(B=-5, \ C=9)$, $(B=-6, \ C=20)$, etc. (I have taken into account the constraint on $B$ to be negative).
Otherwise said,
$$\frac14x^2-4x=0, \ \ \ \frac14x^2-5x+9=0, \ \ \ \frac14x^2-6x+2\sqrt{5}=0$$
have all the same $A$ and the same $\Delta$ without having the same $B$ and the same $C$.