Odpowiedź:
a) [tex]4 {x}^{2} - 1 > 0 \\ a = 4 \: \: \: b = 0 \: \: \: c = - 1 \\ \Delta = {b}^{2} - 4ac \\ \Delta = {0}^{2} - 4 \times 4 \times ( - 1) \\ \Delta = 16 \\ x _{1} = \frac{ - 0- \sqrt{16} }{4 \times 2} \\ x _{1} = \frac{ - 4}{8} \\ x _{1} = - \frac{1}{2} \\ x _{2} = \frac{ - 0 + \sqrt{16} }{4 \times 2} \\ x _{2} = \frac{4}{8} \\ x _{2} = \frac{1}{2} \\ x \in ( - \infty ; - \frac{1}{2} ) \: \cup \: ( \frac{1}{2}; \infty )[/tex]
b) [tex] {x}^{2} + 5x > 2 {x}^{2} - 4x \\ - {x}^{2} + 9x > 0 \\ a = - 1 \: \: \: b = 9 \: \: \: c = 0 \\ \Delta = {9}^{2} - 4 \times ( - 1) \times 0 \\ \Delta = 81 \\ x_{1} = \frac{ - 9 - \sqrt{81} }{ - 1 \times 2} \\ x_{1} = \frac{ - 18}{ - 2} \\ x_{1} = 9 \\ x_{2} = \frac{ - 9 + \sqrt{81} }{ - 1 \times 2} \\ x_{2} = \frac{0}{ - 2} \\ x_{2} = 0 \\ x \in (0;9)[/tex]
