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 \(x < \frac{1}{3}, x > -1\)
waec 2011furthermathematics

The correct answer is B. \(x > -1, x > -\frac{1}{3}\)

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

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

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

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

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

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

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

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

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