(1) In the quadratic equation of one variable: the square of X + BX + C = 0, if the coefficients B and C can be taken in 1,2,3,4,5,6, then the number of equations with real number solutions is [?] (2) If x satisfies the square of X - 3x + 1 = 0, then the value of X + (1 / x) is [?]

(1) In the quadratic equation of one variable: the square of X + BX + C = 0, if the coefficients B and C can be taken in 1,2,3,4,5,6, then the number of equations with real number solutions is [?] (2) If x satisfies the square of X - 3x + 1 = 0, then the value of X + (1 / x) is [?]

First question, first formula, [x ^ 2 + BX + (B / 2) ^ 2] - (B / 2) ^ 2 + C = 0
Get (x + B / 2) ^ 2 = (B / 2) ^ 2-c
Since (x + B / 2) ^ 2 is a complete square form, the complete square form must be greater than or equal to 0. Therefore, if this form is to be solved, there must be (x + B / 2) ^ 2 = (B / 2) ^ 2-C > = 0
That is B ^ 2 > = 4C
When B is 1, C has no solution, so any number of 1, 2, 3, 4, 5, 6 times 4 will not be less than or equal to 1;
Similarly, when B is 2, C can be 1;
When B is 3, C can take 1,2, two sets of solutions;
When B is taken as 4, C can take 1,2,3,4;
When B is 5, C can take 1,2,3,4,5,6 solutions;
When B is 6, C can take 1,2,3,4,5,6 solutions;
There are 1 + 2 + 4 + 6 + 6 = 19 solutions
Question 2: the original formula is x ^ 2 + 1 = 3x
Then x + 1 / x = x ^ 2 / x + 1 / x = (x ^ 2 + 1) / X
Because x ^ 2 + 1 = 3x
So (x ^ 2 + 1) / x = 3x / x = 3