Question

Очень жизненно небходимо решить квадратные уравнения : 1) 5x²-9x-2=0
2)2x²-11x+5=0
3)5x²+11x+2=0
4)3x²-7x+2=0
5)2x²-3x-2=0

Answer 1

$1) \: \: \:5 x {}^{2} - 9x - 2 = 0 \\ d = ( - 9) {}^{2} - 4 \times 5 \times ( - 2) = 81 + 40 = 121 \\ x1 = \frac{9 + 11}{2 \times 5} = \frac{20}{10} = 2 \\ x2 = \frac{9 - 11}{2 \times 5} = - \frac{2}{10} = - \frac{1}{5}$

$2) \: \: \: \: \: 2x {}^{2} - 11x + 5 = 0 \\ d = ( - 11) {}^{2} - 4 \times 2 \times 5 = 121 - 40 = 81 \\ x1 = \frac{11 + 9}{2 \times 2} = \frac{20}{4} = 5 \\ x2 = \frac{11 - 9}{2 \times 2} = \frac{2}{4} = \frac{1}{2}$

Объяснение:

$3) \: \: \: 5x {}^{2} + 11x + 2 =0 \\ d = 11 {}^{2} - 4 \times 5 \times 2 = 121 - 40 = 81 \\ x1 = \frac{ - 11 + 9}{2 \times 5} = - \frac{2}{10 } = - \frac{1}{5} \\ x2 = \frac{ - 11 - 9}{2 \times 5} = - \frac{20}{10} = - 2$

$4) \: \: \: \: 3x {}^{2} - 7x + 2 = 0 \\ d = ( - 7) {}^{2} - 4 \times 3 \times 2 = 49 - 24 = 25 \\ x1 = \frac{7 + 5}{2 \times 3} = \frac{12}{6} = 2 \\ x2 = \frac{7 - 5}{2 \times 3} = \frac{2}{6} = \frac{1}{3}$

$5) \: \: \: \: 2x {}^{2} - 3x - 2 = 0 \\ d = ( - 3) {}^{2} - 4 \times 2 \times ( - 2) = 9 + 16 = 25 \\ x1 = \frac{3 + 5}{2 \times 2} = \frac{8}{4} = 2 \\ x2 = \frac{3 - 5}{2 \times 2} = - \frac{2}{4} = - \frac{1}{2}$

Answer 2

1)5х2−9−2=0,  =5,  =−9,  =−2.

D=b²-4ac=81+40=121=11²

$x_{1} =\frac{-b-\sqrt{D} }{2a} =\frac{9-11}{2*5} =\frac{-2}{10} =-0.2\\$

$x_{2}=\frac{-b+\sqrt{D} }{2a} =\frac{9+11}{10} =\frac{20}{10} =2$

2)2x²-11x+5=0    a=2,  b=-11,   c=5

D=b²-4ac=121-40=81=9²

$x_{1} =\frac{-b-\sqrt{D} }{2a} =\frac{11-9}{4} =\frac{2}{4}=\frac{1}{2} =0.5\\x_{2} =\frac{-b+\sqrt{D} }{2a} =\frac{11+9}{4} =\frac{20}{4} =5$

3)5x²+11x+2=0

D=121-40=81=9²

$x_{1} =\frac{-11-9}{10} =\frac{-20}{10} =-2\\x_{2} =\frac{-11+9}{10} =\frac{-2}{10} =-0.5$

4)3x²-7x+2=0

D=49-24=25=5²

$x_{1} =\frac{7-5}{6} =\frac{2}{6} =\frac{1}{3} \\x_{2} =\frac{7+5}{6} =\frac{12}{6} =2$

5)2x²-3x-2=0

D=9+16=25=5²

$x_{1} =\frac{3-5}{4}=\frac{2}{4} =-\frac{1}{2} =-0.5\\x_{2} =\frac{3+5}{4} =\frac{8}{4} =2$

Удачи!