Is $x^{\frac{1}{2}}+ 2x+3=0$ a quadratic equation Is $$x^{\frac{1}{2}}+ 2x+3=0$$ considered a quadratic equation?
Should the equation be in the form $$ax^2+bx+c=0$$ to be considered quadratic?
 A: Yes, but the quadratic is not in x.
$ 2 (\sqrt x )^2 + (\sqrt x ) + 3 = 0$ can be considered to be quadratic equation in $(\sqrt x ).$
A: if you substitute $\sqrt{x}=u$
so $$ \sqrt{x}+2x+3=u+2u^2+3$$
A: Equations need equal signs, just like sentences need verbs. Just as the equation $ax^2 + bx + c = 0$ is quadratic in $x$, the equation $x^{1/2} + 2x + 3 = 0$ is quadratic in $x^{1/2}$ since it can be written in the form $2(x^{1/2})^2 + x^{1/2} + 3 = 0.$
A: That would not be considered quadratic in $x$, but you can let $u=\sqrt{x}$ to get $2u^2+u+3$. It WOULD be quadratic in $u$. 
A: I think the phrase "quadratic in $x$" is short for "a quadratic polynomial in $x$". This is not a polynomial, so it is not a quadratic polynomial.
However, note that from this equation we can derive another equation that is quadratic in $x$:
$$ x^{1/2} = - (2x + 3)$$
$$x = (2x+3)^2 $$
$$ 4x^2 + 11 x + 9 =0 $$
A: In order for an equation to be quadratic, it needs to be in either the standard form $$ax^2 + bx + c = 0$$, the vertex form (which is easy when dealing with its parabola): $$a(x - h)^2 + k = 0$$ where $h$ is the x-coordinate of the vertex and $k$ is the y-coordinate of the vertex; the factored form $$a(x - h)(x - k) = 0$$, or the intercept form $$a(x - p)(x - q) = 0$$ where $p$ is the x-coordinate of an intercept and $q$ is the y-coordinate of the intercept. Your first equation isn't quadratic, since the degree is $\frac 12$, which is represented as $\sqrt x$ (as some people said already). The equation has to have a degree of 2 in order for it to be quadratic.
A: The equation $f(x)=2x+\sqrt x+3$ is not a quadratic equation since there is no power of $x$ that is $2$ and not all of the powers of $x$ are integers (the square root term).
If we make the transformation $u=\sqrt x$, we get $f(u)=2u^2+u+3$ which is a quadratic equation since the powers are integers and it is of degree $2$.
However, there is an issue in doing this substitution in that extraneous solutions are found (due to plus/minus of square roots).
