I have the quadratic equation $36x^2+80x-35=s^2$

is there any way to make it on the form $AX^2+BX+C^2=s^2$ where $c^2$ is any perfect square.

$s^2$ is unkown but we know it is a perfect square .

Hint: the second equation is an equivalent to the first one so the solution depends on quadratic reduction as i think .

Thanks everybody,


closed as unclear what you're asking by Namaste, Leucippus, steven gregory, hardmath, C. Falcon Apr 15 '17 at 1:43

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ Equivalent in what way? $\endgroup$ – DHMO Apr 14 '17 at 16:47
  • $\begingroup$ that $36x^2+80x−35=s^2$ the other one should equal to $s^2$ too $\endgroup$ – Sherif Abdalmoniem Apr 14 '17 at 16:51
  • $\begingroup$ yes that's what i ment $\endgroup$ – Sherif Abdalmoniem Apr 14 '17 at 16:52
  • $\begingroup$ But you've given an expression, not an equation - do you mean to write $36x^2+80x-35=0$? $\endgroup$ – mrnovice Apr 14 '17 at 16:52
  • $\begingroup$ If your constant term in standard is not already a perfect square, they're is no way to rewrite the expression in standard form so that your constant term is a perfect square. $\endgroup$ – Michael McGovern Apr 14 '17 at 16:53


If you set $x=1$, then the left hand side becomes equal to 81, which is a perfect square. Lets therefore substitute $x=X+1$ into the equation.

$$36(X+1)^2+80(X+1)-35=s^2$$ $$36X^2+72X+36+80X+80-35=s^2$$ $$36X^2+152X+81=s^2$$

This has made the constant term 81, because setting X=0 corresponds to setting x=1 in the original equation and must make the left hand side equal to 81 just as before.


Given that one of the tags is (diophantine-equations), by perfect square I'm guessing you mean the square of an integer.

But if you don't mean that then $-35$ is perfect square of $\sqrt{35}i$.

Beyond that anything you do to the LHS in an attempt to get a perfect square for the constant term will affect the RHS, such that it is probably no longer a perfect square. For example just to convert $-35$ to positive (so the square root is not complex) you either need to multiply by $-1$, which would mean the square root of $-s^2$ is complex. Or you need to add at least 35 to each side. However without knowing what $s$ equals you don't know if $s^2+35$ (taking 35 as an example) is a perfect square.

But you can solve this equation with rational roots. Set $s=13$ then the roots of the equation are $\frac{4}{3}, \frac{-32}{9}$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.