Problem while solving question related to quadratic equations by some another method

Question

Consider following quadratic equation whose roots are $a$ and $b$.

$$x^2 +ax +b = 0 $$

We need to find $a$ and $b$ provided that $a,b\neq0$

So it seems a pretty easy question and it is,

From product of roots we get $a=1$ From sum of roots we get $b=-2$

Now I decided solve with an other way as follows-

Since $a$ and $b$ are roots of equation, therefore they will satisfy it as follows-

$ a^2 + a^2 + b = 0$ and hence $2a^2=-b$

$b^2 + ab +b=0$

Substituting value of $b$ we get

$4a^4-2a^3-2a^2=0$

Since $a\neq0$ solving for a gives two values $a=1$ and $a=-1/2$

and therefore corresponding values of $b$ as $-2,-1/2$ and hence we get following two quadratic equations

$x^2+x-2=0$ and $2x^2-x-1=0$

The first one has roots $(1,-2)=(a,b)$ , therefore we got one answer

but second one has roots $(1,-1/2)\neq(a,b)$

Why does this happens with second equations that it's roots are not equal to $a,b$

"therefore they will satisfy..." This step is not reversible, so the solutions must be verified each. — dxiv, May 24 '21 at 06:34
@lalittolani It's a one way implication, not an equivalence, and so can introduce extraneous solutions. In this case, you are losing the fact that $a,b$ must not be the same root of the original equation. — dxiv, May 24 '21 at 06:36
Information seems to be missing: "From product of roots we get $a = 1$" : How? All I see, prior to that statement is $x^2 + ax + b = 0$. How do you conclude from that equation that $a=1$? — user2661923, May 24 '21 at 06:39
@DatBoi Perhaps, but combining off-beat analysis with missing info is too much. OP needs to respond directly, especially by editing the question. — user2661923, May 24 '21 at 06:41
Information seems to be missing: "From product of roots we get $a=1$" : How? All I see, prior to that statement is $x^2 + ax + b = 0$. How do you conclude from that equation that $a=1$? — user2661923, May 24 '21 at 06:54
everyone knows product of roots = constant term / highest coefficient so if I have mentioned roots and equation clearly and used the term " product of roots" there must be no problems — Lalit Tolani, May 24 '21 at 06:58

score 2 · Accepted Answer · answered May 24 '21 at 07:19

Since $a$ and $b$ are roots of equation, therefore they will satisfy it as follows-

$a^2 + a^2 + b = 0$

$b^2 + ab +b=0$

This system of equations does indeed follow from the problem statement, but it is not equivalent to it. Reason is that the problem assumes $a,b$ to be the roots of the quadratic, with emphasis on the plural of "roots".

The first solution based on Vieta's formulas preserves this assumption, and provides the correct result.
But the substitution that leads to the system of equations loses the distinction that $a,b$ must be the two roots of the quadratic, and allows for $a,b$ to be one and the same root.

In other words, extraneous solutions may be introduced among the solutions of the system, where $a$ is a root and $b=a$. Which is precisely what happened in this case with $a = b = -1/2$.

Therefore, the solutions to the system are not guaranteed to satisfy the original problem, and must be verified each. (Note: it is not safe to automatically reject any solution with $a=b$, because they could be equal in case a real solution existed which turned out to be a quadratic with a double root.)

thankyou for solving my problem fabulously in both your answers — Lalit Tolani, May 25 '21 at 01:59
@lalittolani Yours was a well asked question. It's good that you tried it, and even better that you noticed the inconsistency. The fallacy here was of a more subtle variety than the usual algebraic pitfalls. — dxiv, May 25 '21 at 02:20

Problem while solving question related to quadratic equations by some another method

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