The solutions of the quadratic equation $ax^2+bx+c=0$ are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. If we take the coefficient of x i.e $b$ as zero in the equation and also in solution, then we get the same value of x in both cases but If we take $a$ as zero then we don't get same value of x. Is there any way to get the result from the solution when the coefficient of highest order term is zero?
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When $a=0$ the initial equation becomes $bx+c=0$ with the solution $x=-c/b$. Obviously if you try to get $a=0$ in the equations for the roots you get $0/0$. However, you can approximate in the limit of small $a$: $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{b\left(-1\pm\sqrt{1-\frac{4ac}{b^2}}\right)}{2a}\approx\frac{b}{2a}\left(-1\pm\left(1-\frac{2ac}{b^2}\right)\right)$$ The solution with "-" will go to $\pm\infty$, depending on the sign of $b$. But the solution with "+" is going to be $$\frac{b}{2a}\frac{-2ac}{b^2}=-\frac{c}{b}$$
Andrei
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Another possibility is using the formula $\frac{2c}{-b\pm \sqrt{b^2-4ac}}$ instead. You get that formula if you set $x=1/y$ and solve the quadratic equation obtained by clearing denominators in $a\frac{1}{y^2}+b\frac1y + c =0$.
See also the answers to Numerically stable algorithm for solving the quadratic equation when $a$ is very small or $0$.
Federico Poloni
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If a=0 ,the quadratic formula is meaningless because of the a=0 in the denominator . And the equation is not quadratic ,it is linear .
user439545
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