On the problem of quadratic equation of one variable 1. The univariate quadratic equation (a + C) x ^ 2 + BX - (2c-a) = 0 The sum of the two roots is - 1, and the difference is 1 (1) Find the two roots of this equation; (2) find the value of a: B: C PS: "x ^ 2" means the square of X, the same below 2. It is known that the quadratic equation AX ^ 2 + BX + C = 0 about X. A misinterprets the quadratic coefficients and obtains two of them as 1 and 4 by mistake. B misinterprets the symbols of the coefficient of the first term and the constant term and obtains two of them as - 2 and 6 by mistake and requests the original equation Thank you very much. It's better to have a process, but you can give me some ideas,

On the problem of quadratic equation of one variable 1. The univariate quadratic equation (a + C) x ^ 2 + BX - (2c-a) = 0 The sum of the two roots is - 1, and the difference is 1 (1) Find the two roots of this equation; (2) find the value of a: B: C PS: "x ^ 2" means the square of X, the same below 2. It is known that the quadratic equation AX ^ 2 + BX + C = 0 about X. A misinterprets the quadratic coefficients and obtains two of them as 1 and 4 by mistake. B misinterprets the symbols of the coefficient of the first term and the constant term and obtains two of them as - 2 and 6 by mistake and requests the original equation Thank you very much. It's better to have a process, but you can give me some ideas,

Idea: This is the application of Weida's theorem. 1, (1) let two be x1, x2x1 + x2 = - 1x1-x2 = 1, the solution is: X1 = 0, X2 = - 1 (2) X1 + x2 = - B / (a + C) = - 1x1 * x2 = (a-2c) / (a + C) = 0, the solution is: a = 2CB = 3C, so a: B: C = 2:3:12, because a only misread the coefficient of the quadratic term