Can we compare an equation like I did. Given that $p(x)$ has real roots. If it true, Would it also be true when roots are imaginary? Is it true. Please help
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Try what you did explicitly with any quadratic with $a \geq 2$. – fixedp Dec 16 '16 at 14:30
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Yes I do not get them equal but how to prove ot wrong? I mean what arguement can we give? – user399511 Dec 16 '16 at 14:37
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See the Bifactor Theorem. for a careful proof of the correct result. – Bill Dubuque Dec 16 '16 at 15:23
4 Answers
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It's not quite right. Saying $p(x)$ has two roots $\alpha$ and $\beta$ does not mean that $p(x)=(x-\alpha)(x-\beta)$. You are forgetting that any constant times $(x-\alpha)(x-\beta)$ will still have roots $\alpha$ and $\beta$. So, what you can conclude is that $$ p(x)=c(x-\alpha)(x-\beta), $$ where $c$ is a constant. From the expansion, it follows, in fact, that $c=a$.
ervx
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If a polynomial $p$ of degree $2$ has the roots $\alpha$ and $\beta$, then
$$p(x)=a(x- \alpha)(x - \beta)$$
Fred
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The comparison you did is true, but...
In your case the fact that $a=1$ is wrong, since there are possibilities that the $\alpha$ or $\beta$ may be in fraction.
Harsh Kumar
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Nothing allows you to say that I and II are identical, and they aren't !
All you know is that they have the same roots, i.e. that they cancel at $x=\alpha$ and $x=\beta$, meaning that
$$a\alpha^2+b\alpha+c=a\beta^2+b\beta+c=0,$$
which can be written
$$a(\alpha^2-\beta^2)+b(\alpha-\beta)=0$$
and even
$$a(\alpha+\beta)+b=0$$ if $\alpha$ and $\beta$ are known to be different.
There is nothing more that you can conclude.