Question

The number of real roots of the equation (2x-3/x-1) + 1 = (6x^2 - x -6/x-1) is:

1) 0
2) 2
3) 4
4) None of these​

Answer 1

The two equations can be written as

x

2

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

and x

2

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

Divide by 2

∴ x

2

(6k+2)+

2

p

x + {3k-1)=0 ...(3)

Comparing (1) and (3), we get r =

2

p

∴ 2r-p = 0.

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Answer 2

Answer:

The two equations can be written as

x

2

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

and x

2

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

Divide by 2

∴ x/2

(6k+2)+  2p

x + {3k-1)=0 ...(3)

Comparing (1) and (3), we get r =  2p

∴ 2r-p = 0.

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