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 Post subject: inequility problem
PostPosted: Thu, 17 May 2012 16:34:32 UTC 
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when the range of x determine by -ax^2+bx+4>=0 is (-1/3)<=x<=4 ;
then a=?,b=?
:(


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 Post subject: Re: inequility problem
PostPosted: Thu, 17 May 2012 16:47:44 UTC 
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anik wrote:
when the range of x determine by -ax^2+bx+4>=0 is (-1/3)<=x<=4 ;
then a=?,b=?
:(


What have you tried? Where is your work?

By the way, it is inequality, not inequility.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: inequility problem
PostPosted: Thu, 17 May 2012 17:02:59 UTC 
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anik wrote:
when the range of x determine by -ax^2+bx+4>=0 is (-1/3)<=x<=4 ;

What do you get IF: -ax^2 + bx + 4 = 4 ?

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I'm just an imagination of your figment...


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