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Solve \(3x^{2} + 4x + 1 > 0\)

Further Mathematics
WAEC 2011

Solve \(3x^{2} + 4x + 1 > 0\)

  • A. \(x < -1, x < -\frac{1}{3}\)
  • B. \(x > -1, x > -\frac{1}{3}\)
  • C. \(x > \frac{1}{3}, x < -1\)
  • D. \(x < \frac{1}{3}, x > -1\)
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Correct Answer: Option B
Explanation

\(3x^{2} + 4x + 1 > 0 \)

\(3x^{2} + 3x + x + 1 > 0\)

\(3x(x + 1) + 1(x + 1) > 0\)

\((3x + 1)(x + 1) > 0\)

\(3x + 1 > 0 \implies 3x > -1 \)

\(x > -\frac{1}{3}\)

\(x + 1 > 0 \implies x > -1\)

\(x > -1, x > -\frac{1}{3}\)


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