Question

1.
If the equations k (6x2 + 3) + rx + 2x2 - 1 = 0 and 6k (2x2 + 1) + px + 4x - 2 = 0
common, then the value of (2r-p) is
(A) o
(B) 1/2
(C) 1
(D) none of these
all possible real values of 2 is​

Answer 1

Answer:

0

Step-by-step explanation:

The two equations can be written as

                x2 (6k+2) + rx +(3k-1) = 0                  ...(1)

and         x2 {12k + 4) + px + (6k - 2) = 0.              ...(2)

Divide by 2

∴      x2 (6k+2)+2px + {3k-1)=0                       ...(3)

Comparing (1) and (3), we get r = 2p  ∴    2r-p = 0.