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. \(x< \frac{1}{3}, x > -1\)
Correct Answer: B. \(x > -1, x > -\frac{1}{3}\)
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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