If x2 + kx + \(\frac{16}{9}\) is a perfect square, find the value of k

A.

\(\frac{8}{3}\)

B.

\(\frac{7}{3}\)

C.

\(\frac{5}{3}\)

D.

\(\frac{2}{3}\)

Correct answer is A

x2 + kx + \(\frac{16}{9}\); Perfect square

But, b2 - 4ac = 0, for a perfect square

where a - 1; b = k; c = \(\frac{16}{9}\)

k2 - 4(1) x \(\frac{16}{9}\) = 0

k2 - \(\frac{64}{9}\) = 0

k2 = \(\frac{64}{9}\)

k = \(\sqrt{\frac{64}{9}}\)

k = \(\frac{8}{3}\)