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If \(\begin{vmatrix}  k & k \\ 4 & k \end{vmatr...

If |kk4k|+|231k|=6, find the value of the constant k, where k > 0.

A.

1

B.

2

C.

3

D.

4

Correct answer is C

|kk4k|+|231k|=6

|kk4k|=(k24k)

|231k|=(2k+3)

\therefore (k^{2} - 4k) + (2k + 3) = k^{2} - 2k + 3 = 6

k^{2} - 2k - 3 = 0, factorising, we have k + 1 = 0 or k - 3 = 0

Since k > 0, k = 3.