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

Mathematics
WAEC 2010

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. \(\frac{2}{3}\)
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Correct Answer: Option A
Explanation

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


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