If x 2 + kx + \(\frac{16}{9}\) is a perfect square, find the value of
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. \(\frac{2}{3}\)
Correct Answer: A. \(\frac{8}{3}\)
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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