Question

20 задание
спс зарание​

Answer 1

$1) 2x^{2} = 3x + 2\\2x^{2} - 3x - 2 = 0\\D = b^{2} - 4ac = 9 + 16 = 25\\x_{12} = \frac{-b +-\sqrt{D} }{2a} \\x_{1} = \frac{-b + \sqrt{D} }{2a} = \frac{3 + 5}{4} = 2\\x_{2} = \frac{-b - \sqrt{D} }{2a} = \frac{3-5}{4} = -0.5$

$y_{1} = 2 * 2^{2} =8$

$y_{2}= 2 * (-0.5)^{2} = 0.5$

$2) -0.5x^{2} = 0.5x -3\\0.5x^{2} + 0.5x - 3 = 0\\D = b^{2} - 4ac = 0.25 + 6 \\\\x_{12} = \frac{-b+-\sqrt{D} }{2a} \\x_{1} = \frac{-b + \sqrt{D} }{2a} = \frac{-0.5 + 2.5}{1} = 2\\x_{2} = \frac{-b - \sqrt{D} }{2a} = \frac{-0.5 - 2.5}{1} = -3$

$y_{1} = 0.5 * 4 = -2$

$y_{2} =- 0.5 * (-3)^{2} =- 4.5$

Answer 2

$1)y=2x^{2};y=3x+2\\2x^{2}=3x+2;\\2x^{2}-3x-2=0;\\D= 9+16=25\\x_{1}=\frac{3+5}{4}=2;\\x_{2}=\frac{3-5}{4}=-\frac{1}{2};\\y_{1}=2*4=8;\\y_{2}=2*\frac{1}{4}=0.5;$

Ответ: (2;8), (0.5;0.5)

$2)y=-\frac{1}{2} x^{2};y=\frac{1}{2} x-3;\\-\frac{1}{2}x^{2}=\frac{1}{2}x-3;|*2\\ -x^{2}=x-6;\\-x^{2}-x+6=0;|*(-1)\\x^{2}+x-6=0;\\D = 1 +24=25;\\x_{1}=\frac{-1+5}{2}=2;\\x_{2}=\frac{-1-5}{2}=-3;\\y_{1}=-\frac{1}{2}*4=-2;\\ y_{2}=-\frac{1}{2}*9=-4.5;$

Ответ: (2;-2), (-3;-4.5)