# How are we to correct this error?(from Gelfand's Algebra)

$ax^2+bx+c=0$. If $a=0$,the equation takes the form $bx+c=0$ which has a unique solution $x=-\frac{c}{b}$. Strictly speaking,the last sentence is wrong; when $b=0$ the quotient $\frac cb$ is undefined.How are we to correct this error?

The problem is copied verbatim from Gelfand's algebra text. I do not fully understand the question myself.

If c=0, x has infinite solutions(since any x can be substituted in the equation),otherwise it has none{if c is constant and no x will satisfy the equation(unless it is possible that x=0.Is it?)}. Is this an acceptable answer for the problem?

• If $c=0, x$ has at most two solutions. – Extremal Apr 3 '15 at 20:44
• What is your exact question? Do you have to classify the solutions of $ax^2+bx+c=0$? – Extremal Apr 3 '15 at 20:45
• How? if all coefficients are 0, won't any x satisfy the equation? what would be the 2 solutions? – Aditya Chintalapati Apr 3 '15 at 20:46
• It is important to clearly mention your exact question and the cases you have tried so far. If your case is $a=b=c=0$, then the answer is trivial. – Extremal Apr 3 '15 at 20:47
• The question in the tile is copied verbatim from Gelfand's algebra. I do not fully understand how to address the problem. – Aditya Chintalapati Apr 3 '15 at 20:48

Namely if $b=0,$ can we make the assertion that $bx+c=0 \implies x = -c/b$? As graydad pointed out above, you cannot and instead must treat the two cases individually.
A quadratic can only ever have at most two solutions. In general, a polynomial of degree $n$ $$a_nx^n+a_{n-1}x^{n-1} + \ldots + a_1x+a_0$$ where $a_n \neq 0$ can have at most $n$ distinct roots. If $a = 0$ then you are left with $bx+c=0$. I'm not sure exactly what you are looking for when you say "correct the error" but I would lean toward making cases.
Case 1: $b \neq 0$. Then $bx+c = 0 \implies x = -\frac{c}{b}$. This equality holds no matter what $c$ is.
Case 2: $b = 0$. Then you are left with $c=0$.
• Actually, every non-zero polynomial of degree $n$ has at most $n$ distinct roots, and $0$ is the only polynomial which has infinitely many roots. Also, $c = 0$ is not a root: if $b=0$ then the polynomial is constant, so either it has no roots ($c \neq 0$) or it has infinitely many ($c = 0$). – A.P. Apr 3 '15 at 22:49