The two equations can be written as

x^2 (6k+2) + rx + (3k-1) = 0 …… (i)

& x^2(12k + 4) + px + (6k - 2) = 0 ….. (ii)

Divide (ii) by 2

Thus, we get x^2 (6k+2)+ p/2*x + (3k-1)=0 …. (iii)

Comparing (i) and (iii), we get r = p/2

Thus 2r-p = 2(p/2)-p= 0. (OPTION A)

## If the equations k(6x2+3)+rx+2x2−1=0k(6x2+3)+rx+2x2−1=0 and 6k(2x2+1)+px+4x2−2=06k(2x2+1)+px+4x2−2=0 have both roots common then the value of (2r−p)(2r−p) is Print

Modified on: Tue, 12 Oct, 2021 at 1:03 PM

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