WAEC - Further Mathematics (2011 - No. 8)
Solve \(3x^{2} + 4x + 1 > 0\)
\(x < -1, x < -\frac{1}{3}\)
\(x > -1, x > -\frac{1}{3}\)
\(x > \frac{1}{3}, x < -1\)
\(x < \frac{1}{3}, x > -1\)
ব্যাখ্যা
\(3x^{2} + 4x + 1 > 0 \)
\(3x^{2} + 3x + x + 1 > 0\)
\(3x(x + 1) + 1(x + 1) > 0\)
\((3x + 1)(x + 1) > 0\)
\(3x + 1 > 0 \implies 3x > -1 \)
\(x > -\frac{1}{3}\)
\(x + 1 > 0 \implies x > -1\)
\(x > -1, x > -\frac{1}{3}\)