The value of `K,` for which the equations `2x^(2)+kx-5=0` and `x^(2)-3x-4=0` may have one not in common is/are

Options

A. `-3, (27)/(4)`
B. `3, (-27)/(4)`
C. `-3, (-27)/(4)`
D. `3, (27)/(4)`

Answer

C

Explanation

Let `alpha` be the common root of the two equations. Then, `2alpha^(2) + k alpha - 5 = 0 and, alpha^(2) - 3alpha -4 = 0`
Solving these two equations, we get `(alpha^(2))/(4k -15k)=(alpha)/(-5+8)=(1)/(-6-k)`
`rArr" "alpha^(2) = (4k + 15)/(k+6) and alpha=(-3)/(k+6)`
`rArr" "((-3)/(k+6))^(2) = (4k + 15)/(k+6)" "[therefore alpha^(2) = (alpha)^(2)]`
`rArr" "(4k + 15) (k + 6) = 9`
`rArr" "4k^(2) + 39k + 81 = 0 rArr k = - 3 or, k = -(27)/(4)`