It is known that the quadratic equation AX2 + BX + C = 0 with respect to X has no real roots. If a misinterprets the quadratic coefficients and obtains two of them as 2 and 4, and B misinterprets the sign of the quadratic coefficients and obtains two of them as - 1 and 4, then the value of 2B + 3CA is () A. 2B. 3C. 5D. -6

It is known that the quadratic equation AX2 + BX + C = 0 with respect to X has no real roots. If a misinterprets the quadratic coefficients and obtains two of them as 2 and 4, and B misinterprets the sign of the quadratic coefficients and obtains two of them as - 1 and 4, then the value of 2B + 3CA is () A. 2B. 3C. 5D. -6

For a: let K (X-2) (x-4) = 0, we get kx2-6kx + 8K = 0; for B: let P (x + 1) (x-4) = 0, we get px2-3px-4p = 0. From these two equations, we can see that no matter how wrong, the constants in the equations of a and B are equal, so 8K = - 4P, that is, P = - 2K, then 2B + 3CA = - 12K + 24K − 2K = - 6

Or about Vader's theorem
Given the square of the equation x-3x-2 = 0, do not understand the equation, use the Weida theorem, find an equation, make its root than one of the known equation 1, 1
I have come up with this question, but there is no specific and correct process,
You made a mistake on the first floor. You can't use the Veda theorem
X1 + x2 = 3, which is absolutely true

Just change the result
X ^ 2-3x-3 + radical 17 (- radical 17) = 0

A problem of Weida theorem``
It is known that a and B are two real roots of the quadratic equation 4kx ^ 2 + 4kx + K + 1 (x ^ 2 is the square of x)
(1) Is there a real number k such that (2a-b) × (a-2b) = - 1.5
(2) Find the integer value of the real number k with the value of (A / b) + (B / a) - 2 as an integer
(3) If k = - 2, find the value of Z = A / b
(1) (2) two steps as long as the answer '' (3) give a process``

1) K = 1.8, but from △ > 0, K is obtained

A problem about Weida theorem
Solve the equation and make a new equation so that the product of two real roots and the sum of two squares are the two roots of the following equation: x ^ 2-3x-10 = 0

x²－3x－10＝0
（x－5）（x＋2）＝0
x1＝5,x2＝－2
Let the equation be X & sup2; + BX + C = 0, and its two roots are x1 ′, X2 ′
x1′·x2′＝－2＝c,x1′²＋x2′²＝5
（x1′＋x2′）²－2x1′x2′＝5
（x1′＋x2′）²＝5＋2x1′x2′＝5＋2×（－2）＝1
∴ x1′＋x2′＝±1＝b
The required equation is X & sup2; ± X-2 = 0