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3 Aug 2012 13:50:55 IST

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1.Let alpha,beta,gamma be the roots of (x-a)(x-b)(x-c)=d,d not equal to 0, then the roots

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1.Let alpha,beta,gamma be the roots of (x-a)(x-b)(x-c)=d,d not equal to 0, then the roots of the equation (x-alpha)(x-beta)(x-gamma)+d=0 are:options:(A) a+1,b+1,c+1 (B) a,b,c (C) a-1,b-1,c-1 (D) a/b,b/c,c/a

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vishal gupta

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3 Aug 2012 17:44:09 IST

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correct option is "b"

Parthib

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3 Aug 2012 18:02:38 IST

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Please tell me how you got the answer !

jagdish singh

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3 Aug 2012 20:12:04 IST

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jagdish singh

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3 Aug 2012 20:12:43 IST

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$\hspace{-16}$Here $\bf{\alpha\;,\beta\;,\gamma}$ are The Roots of the equation $\bf{(x-a)(x-b)(x-c)=d}$\\\\\\ $\bf{\Rightarrow (x-a)(x-b)(x-c)-d=(x-\alpha)(x-\beta)(x-\gamma)=0}$\\\\\\ $\bf{\Rightarrow (x-\alpha)(x-\beta)(x-\gamma)+d=(x-a)(x-b)(x-c)=0}$\\\\\\ So From above Line, $\bf{x=a\;,b\;,c}$ are the Roots of the Equation\\\\\\ $\bf{(x-\alpha)(x-\beta)(x-\gamma)+d=0}$$

Rupali Vaish

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5 Aug 2012 12:04:08 IST

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solve hi nahi ho raha yrr

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