Discriminant of an elliptic curve I have found different discriminants for general Weierstrass elliptic curves: 
$y^2 = x^3 + ax + b$
For example, WolframAlpha state it is $-16(4a^3 + 27b^2)$ on their site http://mathworld.wolfram.com/EllipticDiscriminant.html
However, in books I have read like Washington's Elliptic Curves: Number Theory and Cryptography the discriminant is stated to be $-(4a^3 + 27b^2)$. 
Everywhere I look seems to have either of these options, but no one gives a reason for using it. Is there one correct answer or are they both correct? If this is so, why? 
 A: You know the Weirstrass or canonical form of an elliptic curve is a trinomial $x^3+ax+b$ and a cubic curve is elliptic iff its discriminant is non-zero. For this it is not matter if this discriminant has one of the two forms you mention or even if the discriminant is $4A^3+27B^2$ (Cassels).
The factor $-16$ is purely convenient for some purposes but it is irrelevant indeed and it becomes from  distinct birrational transformations leading a general expression of an elliptic cubic to its canonical form as a trinomial.
Furthermore the Weierstrass function $\wp(z)$ is related to its derivative by the fundamental equation
$$[\wp’(z)]^2=4[\wp(z)]^3+A\wp(z)+B$$
for which some authors use the canonical form $4x^3+ax+b$ instead of $x^3+ax+b$.
On the other hand, in Number Theory, the discriminant of a trinomial $x^n+ax+b$ (think Weierstrass for $n=3$) is defined by $$D(1,x,…x^{n-1})= (-1)^{\frac{n(n-1)}{2}}[n^nb^{n-1}+(-1){n-1}(n-1)^{n-1}a^n]$$ hence for the trinomial $x^3+ax+b$ you get
$$D(1,x,x^2)=(-1)^3[27b^2+4a^3]=-(4a^3+27b^2)$$ in accordance with Washington.
Concerning Wolfram, you can see in the first step of his simplification towards the canonical form he gets, the expression $$y^2=4x^3+b_2x^2+2b_4x+b_6$$ and a “simpler” discriminant for “another” polynomial; he writes “The discriminant depends on the choice of equations, and can change after a change of variables” but he is careful to add “unlike the j-invariant” 
Look at the coefficient $4$ of $x^3$ and note the subsequent step of  change of variables trying to see that he could avoid if he have wished the elimination of this coefficient.
That´s all with my deficient English (I hope it can help you something).
A: Question: "Everywhere I look seems to have either of these options, but no one gives a reason for using it. Is there one correct answer or are they both correct? If this is so, why?"
Answer: Given an equation $F(x,y):=y^2-f(x)$ where $f(x) \in k[x]$ is a polynomial of degree $d\geq 3$, it follows the jacobian ideal $J(F):=(y,f(x),f'(x)) \subseteq k[x,y]$ gives rise to the singular subscheme $C_{sing} \subseteq C:=Spec(k[x,y]/(F)$. And $C_{sing}=\emptyset$ iff $J(F)=(1)$ is the unit ideal. The ideal $J(F)$ is the unit ideal iff $f(x)$ and $f'(x)$ have no common root in $k$ and this is iff $Discr(f(x),f'(x))\neq 0$. In the case of $char(k) \neq 2$ it follows $-16$ is a unit in $k$ hence $−16(4a^3+27b^2)\neq 0$ iff $−(4a^3+27b^2) \neq 0$ in $k$. There is the "Sylvester formula" calculating the discriminant $Discr(f(x),f'(x))$ as a determinant.
Note: In general if $f(x) \in k[x]$ is a polynomial, the discriminant $Discr(f(x),f'(x))$ gives a criterion to test if $f(x)$ has mutiple roots without calculating the roots: There is no general formula for the solutions to the equation $f(x)=0$ when $n:=deg(f(x))\geq 5$. The discriminant test is valid for any $n$.
Note: If you begin with a polynomial $F(x,y):=y^2-f(x)$ with $f(x):=(x^3+ax+b)\in \mathbb{Z}[x,y]$ and reduce the polynomial modulo primes $p$, it follows the discriminant $D(f)=u(a,b)$ which is a polynomial in $a,b$, will tell you for which primes $p$ the curve $V(F_p(x,y)) \subseteq \mathbb{A}^2_{\mathbb{F}_p}$ is regular.
Relationship between discriminants and smoothness of curves
