Given that x = - 1 is a root of the equation 2x ^ 2 + ax-a ^ 2 = 0 about X, then a =? Why can't we use Veda's theorem

Given that x = - 1 is a root of the equation 2x ^ 2 + ax-a ^ 2 = 0 about X, then a =? Why can't we use Veda's theorem

A:
X = - 1 is a root of the equation 2x ^ 2 + ax-a ^ 2 = 0
Substituting: 2-a-a ^ 2 = 0
a^2+a-2=0
(a-1)(a+2)=0
The solution is a = 1 or a = - 2
It's not convenient to use Veda's theorem to calculate

It is known that the square of the equation X-9 (a + b) x + AB-1 = 0, X1 and X2 are the two real roots of this equation
① X 1 is not equal to x 2; 2 x 1 x 2 < AB; 3 x 1 square + x 2 square < A2 + B2
What's wrong with that

The discriminant = 81 (a + b) & #178; - 4 (AB-1) = 81 (a + b) & #178; - 4AB + 4 = 80 (a + b) & #178; + (a + b) & #178; - 4AB + 4 = 80 (a + b) & #178; + (a-b) & #178; + 4 > 0, so 1 is correct
x1x2=ab-1

Let the equation x ^ 2 + (A / x) ^ 2 + 7x + (7a / x) + 2A + 12 = 0 have two equal values, and find the value of A

x^2 + (a/x)^2 + 7x + (7a/x) + 2a +12 =0
x^2 + 2a+ (a/x)^2 + 7x + (7a/x) +12 =0
(x+a/x)²+7(x+a/x)+12=0
(x+a/x+3)(x+a/x+4)=0
∴x²+3x+a=0 x²+4x+a=0
∴a=9/4 a=4

If there is only one value of X, the equation x ^ 2 + A ^ 2 / x ^ 2-7x-7a / x + 2A = 0 about X can be established, and the value of a can be obtained
I just want to ask if a can be 0. If a is 0, then there are two real roots 0,7

This is a quartic formula. If it can be formulated in the form of () ^ 4, then x has a unique solution. If you let a be zero, then it will become a quadratic formula, and there must be two solutions. Let's see briefly that this formula can be formulated as a quadratic polynomial with (x + A / x) as a variable, and this polynomial has a constant term. If you continue to formulate it again, it will become the form of () ^ 4, In this formula process, the value of a is determined. The value of a may be more than one, or it may be a numerical solution. This computer does not have matlab installed, I'm too lazy to calculate, sorry