I saw this question by ATUL MISHRA
Why can a quadratic equation have only 2 roots?
then I thought
$$ax^2+bx+c=0$$has $2$ roots $\alpha,\beta$ such that $$\alpha+\beta=\frac{-b}{a}$$and$$\alpha\beta=\frac{c}{a}$$ couldn't it have two pair of solutions such that both the conditions satisfies
i think it may have same meaning as atul's question had but not sure
please don't mark it duplicate i wanna present my Ideas
let we have two pairs
$$\alpha+\beta=\gamma+\delta$$clearly we have by simmatry$$\alpha=\gamma+m$$and$$\beta=\delta-m$$
so in second equation $$\alpha\beta=\gamma\delta$$
$$\implies (\gamma+m)(\delta-m)=\gamma\delta$$ $$\implies \gamma\delta+(\delta-\gamma)m-m^2=\gamma\delta$$ $$\implies (\delta-\gamma)m=m^2$$ $$\implies m=0 $$or$$ \delta=\gamma+m$$
both the results implies that one and only one pair of roots exists in a quadratic equation
Suppose we have:
$$ \alpha+\beta=\frac{-b}{a}=\gamma+\delta $$
and
$$ \alpha\beta=\frac{c}{a}=\gamma\delta $$
Then, after squaring both sides of the top equation, we have
$$ \alpha^2 + 2\alpha\beta+\beta^2=\gamma^2+2\gamma\delta+\delta^2 $$
Now if we subtract four times the second equation from both sides we get $$ \alpha^2 - 2\alpha\beta+\beta^2=\gamma^2-2\gamma\delta+\delta^2 $$ which we can rewrite as $$ (\alpha-\beta)^2=(\gamma-\delta)^2 $$ which has the two solutions $(\alpha-\beta)=\pm(\gamma-\delta)$.
In the first case: $\alpha-\beta=\gamma-\delta$. But now if we add the equation at the top on each side we get $2\alpha=2\gamma$, so $\alpha=\gamma$.
In the second case: $\alpha-\beta=-\gamma+\delta$. But now if we add the equation at the top on each side, we get $2\alpha=2\delta$, so $\alpha=\delta$.
If follows from the first equation that either $\alpha=\gamma$ and $\beta=\delta$, or $\alpha=\delta$ and $\beta=\gamma$.
