2X + 3-x2 = 0 solve the following equation

2X + 3-x2 = 0 solve the following equation

2x+3-x²=0
x²-2x-3=0
(x-3)(x+1)=0
The solution is x = 3 or x = - 1

We can make the equation x + 1 / X-1 + X-1 / x + 1 + 2x + A + 2 / x2-1 = 0 which has only one real root and the sum of all a's values equal to 0
x+1/x-1+x-1/x+1+2x+a+2/x2-1=0
The result is (2x ^ 2 + 2x + 4 + a) / (x ^ 2-1) = 0
The equation has only one real root,
Then 2x ^ 2 + 2x + 4 + a = 0 has only one real root,
That is, 4-8 (4 + a) = 0
When a = - 7 / 2 and a = - 7 / 2, the root of real number is x = - 1 / 2, which satisfies the condition
That is to say, a can only take one value
So the sum of all a values of only one real root is equal to (- 7 / 2). The correct process is like this, but my answer is wrong. How many points can I get

Each teacher's grading standard is different
If all the procedures except the answers are correct, there should be 8 - 11 points

X1 × X2 is the two parts of the equation x ^ - 3x-2 = 0. How much does x ^ + 3x1 × x2 + 2x ^ equal

X1+X2=3,X1X2=-2
X1^2+3X1×X2+X2^2
=(X1+X2)^2+X1X2
=3^2-2
=7

It is known that X1 is the root of the equation x + log3x = 3, and X2 is the root of the equation x + 3 ^ x = 3. Find the value of X1 + x2

X 1 is the abscissa of the intersection of the curve y = log 3 (x) and the line y = - x + 3
X 2 is the abscissa of the intersection of the curve y = 3 ^ X and the line y = - x + 3
Y = log3 (x) and y = 3 ^ X are reciprocal functions, and the image is symmetric with respect to y = X
And the line y = - x + 3 and y = x intersect at the point (3 / 2,3 / 2)
So, (x1 + x2) / 2 = 3 / 2, X1 + x2 = 3
.