Алгебра: Надо дискриминантом решать

Объяснение:1.2.3.4.5.6....

Надо дискриминантом решать

Ответ:

guzel32211

16.09.2021 20:50

Объяснение:

${x}^{2} - 2x - 15 = 0 \\ d = {2}^{2} - 4 \times ( - 15) = 4 + 60 = 64 \\ \\ x1 = \frac{2 - \sqrt{64} }{2} = \frac{2 - 8}{2} = \frac{ - 6}{2} = - 3 \\ \\ x2 = \frac{2 + \sqrt{64} }{2} = \frac{2 + 8}{2} = \frac{10}{2} = 5$

${2x}^{2} + 3x + 1 = 0 \\ d = {3}^{2} - 4(1 \times 2) = 9 - 8 = 1 \\ \\ x1 = \frac{ - 3 - \sqrt{1} }{2 \times 2} = \frac{ - 3 - 1}{4} = \frac{ - 4}{4} = - 1 \\ \\ x2 = \frac{ - 3 + \sqrt{1} }{2 \times 2} = \frac{ - 3 + 1}{4} = \frac{ - 2}{4} = - 0.5$

${x }^{2} = 2x + 48 \\ {x}^{2} - 2x - 48 = 0 \\ d = 4 - 4 \times ( - 48) = 4 + 192 = 196 \\ \\ x1 = \frac{2 - \sqrt{196} }{2} = \frac{2 - 14}{2} = \frac{ - 12}{2} = - 6 \\ \\ x2 = \frac{2 + \sqrt{196} }{2} = \frac{2 + 14}{2} = \frac{16}{2} = 8$

${x}^{2} - 3x - 18 = 0 \\ d = 9 - 4 \times ( - 18) = 9 + 72 = 81 \\ \\ x1 = \frac{3 - \sqrt{81} }{2} = \frac{3 - 9}{2} = \frac{ - 6}{2} = - 3 \\ \\ x2 = \frac{3 + \sqrt{81} }{2} = \frac{3 + 9}{2} = \frac{12}{2} = 6$

${3x}^{2} + x - 4 = 0 \\ d = 1 - 4 \times ( - 4 \times 3) = 1 + 48 = 49 \\ \\ x1 = \frac{ - 1 - \sqrt{49} }{3 \times 2} = \frac{ - 1 - 7}{6} = \frac{ - 8}{6} = - 1 \frac{1}{3} \\ \\ x2 = \frac{ - 1 + \sqrt{49} }{3 \times 2} = \frac{ - 1 + 7}{6} = \frac{6}{6} = 1$

${x}^{2} = 4x + 96 \\ {x}^{2} - 4x - 96 = 0 \\ d = 16 - 4 \times ( - 96) = 16 + 384 = 400 \\ \\ x1 = \frac{4 - \sqrt{400} }{2} = \frac{4 - 20}{2} = \frac{ - 16}{2} = - 8 \\ \\ x2 = \frac{4 + \sqrt{400} }{2} = \frac{4 + 20}{2} = \frac{24}{2} = 12$