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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: quadratic iitjee question
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[Post New]14 Apr 2008 21:22:33 IST
    Subject: quadratic iitjee question
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hhitesh_1 (79)

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if x^2 + (a-b)x + (1-a-b) = 0 where a,b are real numbers....then find the values of 'a' for which the equation has unequal real roots  for all values of b
    
[Post New]14 Apr 2008 21:51:04 IST
    Subject: Re:quadratic iitjee question
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sboosy (2970)

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x^2+(a-b)x+1-a-b = 0 \\ \\ \mbox{Since the roots are real and distinct} \ D>0 \\ \\ \Rightarrow (a-b)^2-4(1-a-b)>0 \\ \\ \Rightarrow b^2+(4-2a)b+a^2+4a-4>0 \\ \\ \mbox{Quadratic with coeff of square term positive...always} \ > 0 \ \ \Rightarrow D<0 \\ \\ \Rightarrow (4-2a)^2-4(a^2+4a-4)<0 \\ \\ \Rightarrow 32a>32 \\ \\ \Rightarrow a>1
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[Post New]15 Apr 2008 16:03:41 IST
    Subject: quadratic iitjee question
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hhitesh_1 (79)

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arey yaar it was so easy....main all values of b ke chakkar mein sochne lag gaya tha......so stupid of me......dimaag hi nahi chala
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