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Question

If 7Cr+37Cr+1+37Cr+2+7Cr+3>10C4, then the quadratic equation whose roots are α,β and αr1,βr1 have

A
no common roots
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B
only one common root
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C
two common root
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D
none of these
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Solution

The correct option is C two common root
7Cr+37Cr+1+37Cr+2+7Cr+3>10C4
7Cr+7Cr+1+2(7Cr+1+7Cr+2)+7Cr+2+7Cr+3>10C4 8Cr+1+28Cr+2+8Cr+3>10C4
10Cr+3>10C4
r+3=5r=2
Now roots of two different equation are α,β and αr1,βr1
i.e. α1,β1 and α21,β21, i.e. α1,β1 and α1,β1 both roots are same
So Number of common roots are 2.

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