Question

answer the above question​

Answer 1

Explanation:

Putting value of x in equation            2(-k/2)^2 + k - 6(-k/2) -3k =0               2(k^2/4) + k + 3k - 3k = 0                   k^2/2+k=0                                              k^2+2k=0                                              k = 0 or k = -2                                        putting values of k in x LHS = RHS Hence x=-k/2 is a root