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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,

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

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    $\begingroup$ 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
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$$36x^2+80x-35=s^2$$

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.

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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 $\frac{4}{3}, \frac{-32}{9}$.

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