$(a+c)ax^2 + 2bx + c = 2(ac-b^2)(x^2+1)$
If I bring the RHS to LHS and then expand , it will take a lot of time for me to first expand and then find the discriminant for this equation whence is the first way to find the discriminant. Is there a quicker way with which I can find its discriminant ? Maybe just by looking at the equation , not even have to expand.
Please do help me find an approach.
You can tell at a glance that the RHS has no linear term in $x$. You can also tell at a glance that the quadratic coefficient on the RHS is $2ac-2b^2$. The quadratic coefficient on the LHS is $a^2+ca$, and if you subtract the LHS from the RHS, the resulting quadratic coefficient is $a^2+2b^2-ac$. The linear coefficient is just what it is on the LHS, since the RHS has no linear term, so that is $2b$. The constant term is also told at a glance; it is $c$ on the LHS and $2ac-2b^2$.
The discriminant is $y^2-4xz$, where $x,y,z$ are the quadratic, linear and constant coefficients respectively. $x=a^2+2b^2-ac$, $y=2b$, $z=c+2b^2-2ac$. Slightly messily, we compute the discriminant as:
$$\begin{align}4b^2-4(a^2+2b^2-ac)(c+2b^2-2ac)&=4b^2-4(a^2c+2a^2b^2-2a^3c+2b^2c+4b^4-2acb^2-ac^2-2acb^2+2a^2c^2)\\&=4b^2-4(4b^4+2a^2b^2+2a^2c^2+a^2c+2b^2c-2a^3c-4acb^2-ac^2)\\&=4b^2+16acb^2+4ac^2+8a^3c-16b^4-8a^2b^2-8a^2c^2-4a^2c-8b^2c\\&=4b^2(1+4ac-4b^2-2a^2-2c)+4a(c^2+2a^2c-2ac^2-ac)\end{align}$$
It was easy to find the coefficients with a glance; not quite so easy to compute the discriminant itself. There may well be mistakes in my expansion, with so many terms.