If the two roots of the equation AX ^ 2 + BX + C = 0 (a is not equal to 0) are 1, - 1 respectively, then 2A + C=

If the two roots of the equation AX ^ 2 + BX + C = 0 (a is not equal to 0) are 1, - 1 respectively, then 2A + C=

It can be seen from Veda's theorem
x1+x2=-b/a
x1x2=c/a
The two roots are 1, - 1
So X1 + x2 = 0
x1x2=-1
be
-b/a=0
c/a=-1
Because a is not equal to 0
So B = 0
c=-a
So 2A + C = 2a-a = a
I wish you progress in your study

The equation AX ^ 2 + (1 + 2I) x-2a (1-I) = 0 has real roots
The value of real number a in complex equation with real root

ax^2+x+2xi-2a+2ai=0,(ax^2+x-2a)+(2x+2a)i=0
So ax ^ 2 + x-2a = 0, 2x + 2A = 0
A = 0 or positive and negative root sign 3

If the equation y square - 2Y + n = 0 about y has two unequal real roots, judge the case of the roots of equation (n-2) y square - 2ny - (n-3) = 0,
emergency

y²- 2y+ n =0
4 - 4N > 0, so n < 1
(n-2)y² -2ny -(n-3) =0
（-2n）² - 4（n-2 ）[-（n-3）]
= 4n² + 4(n²-5n+6)
=8n² -20n +24
=8(n²-5/2n+5/4²)-8 x 25/16 +24
=8(n-5/4)² +23/2 >=23/2 > 0
So the equation has two unequal real roots

The proof of (n-1) x ^ 2 + MX + 1 = 0 has two unequal real roots: the case of the roots of the equation m ^ 2Y ^ 2-2my-m ^ 2-2n ^ 2 + 3 = 0 of Y
Such as the title
M ^ 2Y ^ 2 is the second power of M multiplied by the second power of Y

Solution: the equation (n-1) x ^ 2 + MX + 1 = 0 has two unequal real roots. Obviously, m ≠ 0 △ = m ^ 2-4 (n-1) > 0m ^ 2 > 4n-4, that is, n > 1n-1 > 0m ^ 2Y ^ 2-2my-m ^ 2-2n ^ 2 + 3 = 0 △ = 4m ^ 2-4m ^ 2 * (- m ^ 2-2n ^ 2 + 3) = 4m ^ 2 * (m ^ 2 + 2n ^ 2-2) = 4 * (4n-4) * [4n-4 + 2n ^ 2-2] = 16 (n-1) * [2 (n ^ 2 + 2n-3)