Quadratic Equation

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Prosenjit Basak
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Quadratic Equation

Unread post by Prosenjit Basak » Wed Feb 20, 2013 10:45 pm

Euler said that every quadratic equation has two roots. But check this out
$x^2+2x+1$
This equation has only one root $-1$. Now what happened here? (I was just a little bit confused where to post this kind of problem)
Yesterday is past, tomorrow is a mystery but today is a gift.

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Fahim Shahriar
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Re: Quadratic Equation

Unread post by Fahim Shahriar » Thu Feb 21, 2013 12:15 am

For the equation $ax^2+bx+c=0$, $x = \frac {-b \pm \sqrt {b^2-4ac}}{2a}$

[$M=b^2-4ac$]
When M>0 ; it has two real roots.
When M<0 ; it has two complex roots.
When M=0 ; its two roots become same.
Name: Fahim Shahriar Shakkhor
Notre Dame College

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nayel
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Re: Quadratic Equation

Unread post by nayel » Thu Feb 21, 2013 6:01 am

Your polynomial does have two (not necessarily distinct) roots: $-1$ and $-1$, because it can factored as $(x+1)(x+1)$. In this case $-1$ is called a multiple root.
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

SMMamun
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Re: Quadratic Equation

Unread post by SMMamun » Thu Feb 21, 2013 3:54 pm

Dear Prosenjit, your confusion is not unjustified. The equation $x^2+2x+1=0$ does 'in reality' have one root, $-1$. But then $x+1=0$ or $x^3+3x^2+3x+1=0$ has also only one root each, $-1$. To make things logically consistent and to differentiate among these equations in terms of their roots, we say that $x+1=0$ has only one root while $x^2+2x+1=0$ has two roots, but of same values. This is especially because the modern approach of root determination is intrinsically linked with the factorization of the algebraic expression. You can no more factorize $(x+1)$, but you can factorize $x^3+3x^2+3x+1$ as $(x+1)(x+1)(x+1)$ . Thus, as you realize, if an algebraic expression contains $n$ as the highest power of a specific variable, we can factorize the expression as the product of $n$ number of linear factors (i.e. of power $1$), each of which will generate exactly one root—roots may not necessarily be all distinct.

In a somewhat similar line of reasoning, $x^2=4x$ must give us two roots because we should factorize it for $x$, not cancel $x$ out from each side. But recall also, $\frac{1}{x^2}=\frac{1}{4x}$ is not equivalent to $x^2=4x$ because the former has only one root!

However, in ancient times, when negative numbers were not discovered or were something mysterious, even a quadratic had either only one positive root or two positive roots, but never one positive and one negative. Thus some quadratics had no roots at all. And even further later, until sixteenth century, some other quadratics still had no roots: this time because imaginary numbers were not conceived well by the mathematicians.

Thus your confusion can be dispelled by studying the endeavors of the mathematicians: how they tried to organize mathematics systematically and logically over the ages. And, of course, I appreciate the spirit of your confusion. :)

Prosenjit Basak
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Joined:Wed Nov 28, 2012 12:48 pm

Re: Quadratic Equation

Unread post by Prosenjit Basak » Thu Feb 21, 2013 9:53 pm

Ow.That means this very quadratic equation actually has two roots.But the roots are same? Hmm Interesting.
Yesterday is past, tomorrow is a mystery but today is a gift.

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