Solve \(3x^{2} + 4x + 1 > 0\)
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option B
Explanation:
\(3x^{2} + 4x + 1 > 0 \)
\(3x^{2} + 3x + x + 1 > 0\)
\(3x(x + 1) + 1(x + 1) > 0\)
\((3x + 1)(x + 1) > 0\)
\(3x + 1 > 0 \implies 3x > -1 \)
\(x > -\frac{1}{3}\)
\(x + 1 > 0 \implies x > -1\)
\(x > -1, x > -\frac{1}{3}\)
\(3x^{2} + 4x + 1 > 0 \)
\(3x^{2} + 3x + x + 1 > 0\)
\(3x(x + 1) + 1(x + 1) > 0\)
\((3x + 1)(x + 1) > 0\)
\(3x + 1 > 0 \implies 3x > -1 \)
\(x > -\frac{1}{3}\)
\(x + 1 > 0 \implies x > -1\)
\(x > -1, x > -\frac{1}{3}\)