Limit of the root of quadratic equation

Question

The root of the equation $ a x^2 + bx + c = 0 $ is given by $$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \;\;\;...(1) $$

On the other hand, if $a = 0$, then from the original equation we get $$ x = - \frac{c}{b} \;\;\;...(2) $$

So I am guessing that as $a \to 0$, one of the root in (1) converges to (2), and the other diverges, but I cannot formally prove it.

Can someone give me a proof, or is my guess incorrect?

See this answer for a view more conceptual than rationalizing the numberator. — Bill Dubuque, Jun 23 '16 at 03:40

André Nicolas · Accepted Answer · 2016-06-23T01:47:39.587

5

Hint: The case $c=0$ requires separate treatment. For $c\ne 0$. multiply top and bottom by $-b\mp\sqrt{b^2-4ac}$, do some cancellation, and take the limit as $a\to 0$. Your guess will be confirmed.

Remark: When you do the multiplication and cancellation, you will obtain the Citardauq Formula for the roots of the quadratic.

edited Jun 23 '16 at 01:47

answered Jun 23 '16 at 01:37

André Nicolas

514,336

...Did you really just write the word "quadratic" backwards? – Noble Mushtak Jun 23 '16 at 01:47
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Yes. In my defense, I did not invent the name! The formula can in fact be computationally useful, in cases of catastrophic almost cancellation. – André Nicolas Jun 23 '16 at 01:48
Thank you very much! Very clear! – Kota Mori Jun 23 '16 at 02:05
@KotaMori: You are welcome. – André Nicolas Jun 23 '16 at 02:08

Limit of the root of quadratic equation

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