Question

квадратные уравнения

Answer 1

1)2y + 2y² = 0                                            (a + b)² = a²+ 2ab + b²

 2y(1 + y) = 0                                             (a - b)² = a² - 2ab + b²

 2y = 0 => y = 0                                        D = b² - 4ac

 1 + y = 0 => y = -1                                      x1,2 = (-b ± √D) / 2a

2)8x² + x - 75 = 0

  D = 1 - 4 * 8 * (-75) = 2401

  x1 = (-1 + 49) / (2 * 8) = 48/16 = 3

  x2 = (-1 - 49) / (2 * 8) = -50/16 = -25/8

3) -2x² - 5x - 2 = 0

   D = 25 - 4 * (-2) * (-2) = 9

   x1 = (5 + 3) / (2 * (-2)) = 8/-4 = -2

   x2 = (5 - 3) / (2 * (-2)) = 2/-4 = -0,5

4) (x + 1)² = (2x - 1)²

   x² + 2x + 1 = 4x² - 4x + 1

   -3x² + 6x = 0

   -3x(x - 2) = 0

   -3x = 0 => x = 0

   x - 2 = 0 => x = 2

Answer 2

1) 2у +2у^2 = 0

2у(1+у)=0

2у=0  1+у=0

у=0      у= - 1

2) $8x^{2} + x - 75 = 0$

Д = $b^{2} -4ac$ = 1 + 4*8*75 = 2401 =$49^{2}$

$=frac{-1 frac{+}{-} sqrt{49^{2} } } {2*8}$

x1 = 48/16 = 3  

x2 = 3,125

3) $-2x^{2} -5x-2=0 | *  (-1)\2x^{2} +5x+2=0\D=b^{2} -4ac= 25 - 16=9=3^{2} \\x_{12} =frac{-bfrac{+}{-}sqrt{D}  }{2a} = frac{-5frac{+}{-}sqrt{3^{2} }  }{2*2}\\x_{1} =-frac{1}{2}      \x_{2} =-2$

4)$(x+1)^{2} =(2x-1)^{2} \x^{2} +2x+1=4x^{2} -4x+1\3x^{2} - 5x=0\x(3x-5)=0\x=0    \x=frac{5}{3} =1frac{2}{3}$