Question

Уравнение к виду ах^2+bx+с=0 , а = не равняется 0, и найти его корни. 1) х(х+6)=7 2) х(х+7)=18 3) 2х(х-5)-4=х(4-х)+1 4) 3х(х+3)-2=1-х(2х+5) , нужно)

Answer 1

$1) x(x+6) = 7 \\\\ x^2+6x =7\\\\ x^2+6x-7 = 0\\\\ x_1*x_2 = -7\\\\ x_1+x_2 = -6\\\\ x_1 = -7, x_2 = 1\\\\ 2) x(x+7) = 18\\\\ x^2 + 7x = 18\\\\ x^2+7x-18 = 0\\\\ x_1*x_2 = -18\\\\ x_1+x_2 = -7\\\\ x_1 = -9, x_2 = 2\\\\$

$3) 2x(x - 5) - 4 = x(4-x) + 1\\\\ 2x^2 -10x -4 = 4x - x^2 +1\\\\ 3x^2-14x-5=0\\\\ D = 196 + 60 = 256\\\\ x_1 = (14 - 16)/6 = -2/6 = -1/3\\\\ x_2 = (14 + 16)/6 = 5\\\\ 4) 3x(x + 3) -2 = 1-x(2x+5)\\\\ 3x^2+9x -2 = 1 - 2x^2 -5x\\\\ 5x^2+14x-3=0\\\\ D = 196 + 60 = 256\\\\ x_1 = (-14+16)/10 = 1/5\\\\ x_1 = (-14-16)/10 = -3\\\\$