I am studying physics and end up with a quadratic equation in this form below. It it mentioned in the book that we need to find the discriminant to proceed but does not show how it is done. The book mentioned to solve for $x$ so $y$ is real.
$$ y=ax^2+c_1\pm\sqrt{c_2+c_3x^2}\\ $$
$c_1, c_2, c_3$ are constant and are different from each other.
How to find a discriminant of that equation? is it possible?
any help or lead will be appreciated.
EDIT:
the equation before rearrangement and the problem from the book.
You are solving for $x$ so that $y$ is real.
$y=ax^2+c_1\pm\sqrt{c_2+c_3x^2}$
First of all you have a restiriction on $x$ that $c_2 + c_3x^2 \ge 0$ because a negative value in an even powered radical is not a real result.
So $c_3x^2 \ge -c_2$ and $x^2 \ge -\frac {c_2}{c_3}$ and because $x^2 \ge 0$ we have the restriction that $x^2 \ge \max(0, -\frac {c_2}{c_3})$. We'll keep that in mind. Note: If $\frac {c_2}{c_3} \ge 0$ then this is not an issue.
$y=ax^2+c_1\pm\sqrt{c_2+c_3x^2}$
$y - ax^2 - c_1 = \pm\sqrt{c_2+c_3x^2}$
$(y-ax^2 - c_1)^2 = c_2 +c_3 x^2$ (note: this is why we need the restriction $x^2 \ge \max(0, -\frac {c_2}{c_3})$. By squaring we added extraneous solutions.)
$a^2x^4 - 2a(y-c_1)x^2 + (y-c_1)^2 = c_2 + c_3 x^2$
$a^2x^4 - [2a(y-c_1)- c_3]x^2 +[(y-c_1)^2 - c_2] = 0$
So $x^2 = \frac { [2a(y-c_1)- c_3]\pm \sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}}{2a^2}$.
We have some restrictions. $[2a(y-c_1)- c_3]^2 +4a^2c_2$ must be non-negative so that $4a^2c_2 \ge - [2a(y-c_1)- c_3]^2$. And as $x^2 \ge 0$. We must have $[2a(y-c_1)- c_3] \ge - \sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}$ and if $ [2a(y-c_1)- c_3] < \sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}$ we will not accept $x^2 = \frac { [2a(y-c_1)- c_3]- \sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}}{2a^2}$ as a solution.
So $x = \pm\sqrt{\frac { [2a(y-c_1)- c_3]\pm \sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}}{2a^2}}=\pm \frac{\sqrt{[2a(y-c_1)- c_3]\pm \sqrt{ [2a(y-c_1)- c_3]^2 +4a^2c_2}}}{\sqrt {2} a} $
With the restriction that everything under the radicals are positive (If you have wrong values of $y, c_1,c_2,c_3, a$ it just won't work) and the restriction that $x^2 \ge \max(0, -\frac {c_2}{c_3})$. Or in other words either $x\ge \sqrt{\max(0, -\frac {c_2}{c_3})}$ or $x \le - \sqrt{\max(0, -\frac {c_2}{c_3})}$.
There could be as many as $4$ solutions or as few as $0$ depending on the restrictions.
Note: If $\frac {c_2}{c_3} \ge 0$ then this is not an issue.