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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

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    http://math.stackexchange.com/questions/2049871/why-can-a-quadratic-equation-have-only-2-roots – 2017-01-08
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    i had seen it there but my answer is not a match for that question as he questioned – 2017-01-08

2 Answers 2

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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

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    Your solutions is right but it makes sense if you think about the solution of the system. However, if you think about the solution of the equation, you could ask yourself if would be possible to have more then $2$ roots. I mean, $3$ or $4$ or $5$ roots and not only pairs of solutions. – 2017-01-08
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    for that we have a standard proof given in books – 2017-01-08
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    i posted it to know errors – 2017-01-08
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    thanks for telling me that – 2017-01-08
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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 $\beta=\gamma$.