Solving for $f$ when $f(x + y) = f(x) + f(y) + axy$ where $a$ is a real number Let $f$ be a real valued continuous function such that $f(x + y) = f(x) + f(y) + axy$ where $a$ is a real number.
We have to show that $f(x) = \frac{a}{2}x^2 + bx$ where $b = f(1) - \frac{a}{2}$
Here is my solution so far:
Let $x = y = 0$ . This gives $f(0) = 0$
Now putting $y = -x$ gives $ax^2 = f(x) + f(-x)$. This gives $a = f(1) + f(-1)$
Also by letting $x = y$ we obtain $f(2x) = 2f(x) + ax^2$
After juggling with these equations I am not able to show that $f(x)$ is a quadratic equation.
Any help would be appreciated.

Let $g(x) = f(x) - \frac{a}{2}x^2$.
Then, subtracting $\frac{a}{2}x^2 + \frac{a}{2}y^2$ from both sides of the given expression, we get : 
$f(x+y) - \frac{a}{2}x^2 - \frac{a}{2}y^2 -axy = f(x) - \frac{a}{2}x^2 +f(y) - \frac{a}{2}y^2 $
$f(x+y) - (\frac{a}{2}(x+y))^2 = f(x) - \frac{a}{2}x^2 + f(y) - \frac{a}{2}y^2$
$g(x+y) = g(x) + g(y)$
This equation is known to have the solution $g(x) = bx$. (Cauchy's equation)
Thus, $f(x)$ has the solution of the form $\frac{a}{2}x^2 + bx$.
