The quadratic function:(BOMC)

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sourav das
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The quadratic function:(BOMC)

Unread post by sourav das » Wed Oct 26, 2011 9:39 am

A quadratic function is in a form of
$f(x)=ax^2 + bx + c$
Any problem and discussion in this section can be done here.
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TOWFIQUL
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Re: The quadratic function:(BOMC)

Unread post by TOWFIQUL » Wed Oct 26, 2011 11:21 am

Page no 5 isn't printed in my book..!! :o

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*Mahi*
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Re: The quadratic function:(BOMC)

Unread post by *Mahi* » Wed Oct 26, 2011 12:51 pm

Then get the soft copy of the book.
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sm.joty
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Re: The quadratic function:(BOMC)

Unread post by sm.joty » Wed Oct 26, 2011 3:08 pm

আমরা বইয়ের পৃষ্ঠা ৬ এর শেষে দেখলাম ,
a>0 হলে নিশ্চায়ক \[\Delta < 0\]
কিন্তু Ez-1.16 এ a>0 আবার \[\Delta > 0\]
ব্যাপারটা কি বুঝলাম না?? :?: :?
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sm.joty
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Re: The quadratic function:(BOMC)

Unread post by sm.joty » Wed Oct 26, 2011 3:22 pm

Example 1.2.5: (new book):
\[\Delta =[(a^2-b^2-c^2)-4b^2c^2]<0\]
\[\Leftrightarrow [a+b+c][a-b-c][a-b+c]][a+b-c]<0\]
\[\Leftrightarrow [b+c-a][c+a-b][a+b-c]>0\]
এখানে [a+b+c] কোথায় গেল ? বুঝি নাই...। :? :? :cry:
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FahimFerdous
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Re: The quadratic function:(BOMC)

Unread post by FahimFerdous » Wed Oct 26, 2011 5:49 pm

Joty, as a,b,c>0 then (a+b+c) must be positive. They just divided both sides by (a+b+c), that's it.
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*Mahi*
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Re: The quadratic function:(BOMC)

Unread post by *Mahi* » Wed Oct 26, 2011 5:55 pm

sm.joty wrote:
\[\Delta =[(a^2-b^2-c^2)-4b^2c^2]<0\]
\[\Leftrightarrow [a+b+c][a-b-c][a-b+c]][a+b-c]<0\]
\[\Leftrightarrow [b+c-a][c+a-b][a+b-c]>0\]
$[a+b+c][a-b-c][a-b+c]][a+b-c]<0$ is just divided by $-[a+b+c]$ here.
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rafid
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Re: The quadratic function:(BOMC)

Unread post by rafid » Wed Oct 26, 2011 9:57 pm

so where did the minus sign go?

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Nadim Ul Abrar
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Re: The quadratic function:(BOMC)

Unread post by Nadim Ul Abrar » Wed Oct 26, 2011 10:22 pm

@ রাফিদ ঃ

সহজ বাংলায় মাইনাসে মাইনাসে প্লাস হয়ে গেসে ।।।
ভাল করে -[a+b+c] গুন করার আগে ও পরের তফাতটা দেখ ।।
$\frac{1}{0}$

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nafistiham
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Re: The quadratic function:(BOMC)

Unread post by nafistiham » Wed Oct 26, 2011 10:39 pm

just like this:
$\left [ a-b-c \right ]$ has turned $ \left [ b+c-a \right ]$
:lol:
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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