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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. enter image description here

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  • $\begingroup$ Rearrange as to leave the square root on one side, then square the equation. You'll get a quadratic in $x^2$. $\endgroup$ – dxiv Dec 18 '17 at 18:22
  • $\begingroup$ you will get $y$ in the discriminant then, is that what you meant? $\endgroup$ – Codelearner777 Dec 18 '17 at 18:26
  • $\begingroup$ Right. I assume you want to solve for $x$ since the question says end up with a quadratic equation. $\endgroup$ – dxiv Dec 18 '17 at 18:34
  • $\begingroup$ It's very unclear what you mean. Ar you solving for $x$? Are restricting your domain so that the discriminant (which is $c_2 + c_3x^2$ and right there) is positive? When you say we need to do this to "proceed", to proceed doing what, exactly? Unless I know what our goal is, I'm more than happy to just write down $y = ax^2 + c_1 + \sqrt{c_2 + c_3x^2}$ on a table napkin and go out for coffee. $\endgroup$ – fleablood Dec 18 '17 at 18:43
  • $\begingroup$ Yes, I want to solve for $x$ so $y$ is real. I will get $y$ in my discriminant equation if I follow your suggestion, (I did this kind of rearrangement before for other purpose in this same physics problem i am facing). if so, the discriminant will depend on $y$, how to deal with that? $\endgroup$ – Codelearner777 Dec 18 '17 at 18:45
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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.

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  • $\begingroup$ wow...I did not expect line by line solutions, but thanks anyway, I will read that. $\endgroup$ – Codelearner777 Dec 18 '17 at 19:13
  • $\begingroup$ +1 for what seems to be a very carefully analyzed algebraic discussion. (I haven't checked all the details, however, but I'll go on my gut instinct.) $\endgroup$ – Dave L. Renfro Dec 18 '17 at 19:21

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