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Indian mathematician Shridharacharya and Persian Mathematician Omar Khayyam (may be others also) are known to have found the formula for roots of a quadratic equation $ax^2+bx+c=0$ as $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}~~~~(1)$$ independently by forming perfect square SLIGHTLY differently.

Currently, Dr. Proshen Loh of Carnegie Mellon University

https://www.popularmechanics.com/science/math/a30152083/solve-quadratic-equations/?fbclid=IwAR2la-1SiNTfUbCfce8VUbbXnH2zE7iB2nUd5vWkwcZeWZu8-rv86brrRVw

has come up with a SLIGHTLY different alternative for finding the roots by solving $$(-\frac{b}{2a}+u) (-\frac{b}{2a}-u)=\frac{b^2}{4a^2}-u^2=\frac{c}{a}~~~~(2)$$ So $x_1,x_2=(-\frac{b}{2a}\pm u)$ will be the roots whose sum is $-b/a$ and product is $c/a$.

One more SLIGHTLY different idea will be to remove second (linear) term of the quadratic as in the case of a cubic:$ax^3+bx^2+cx+d=0$ by using $x=u+p$, then we get $$a(u+p)^2+b(u+p)+c=0 \implies au^2+(2ap+b)u+ap^2+bp+c=0~~~~(3)$$ By setting $2ap+b=0 \implies p=-\frac{b}{2a}$ in (3), we again get the equation (2) for $u$.

The question is: can there be another SLIGHTLY different way for finding the roots of a quadratic equation.

Raffaele
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Z Ahmed
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    For quadratic equations with integer coefficients, yes; the rational roots theorem asserts that if $p/q$ is a rational root of a polynomial with integer coefficients in irreducible form, the, $p$ is a divisor of the constant trm and $q$ is a divisor of the leading coefficient. There results from this that for a monic polynomial; rational roots, if any, are integers. – Bernard Dec 19 '20 at 18:39
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    You could perhaps invoke the algebraic identity $(x_1-x_2)^2 = (x_1+x_2)^2 - 4 x_1 x_2$. – Gribouillis Dec 19 '20 at 21:45

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