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=5⇒r=2
Now roots of two
different equation are α,β and
αr−1,βr−1
i.e. α1,β1 and α2−1,β2−1, i.e. α1,β1 and α1,β1 ⇒ both roots are same
So Number of common roots are 2.