1. If two of the univariate quadratic equations 2x ^ 2-mx + 4 = 0 are x1, X2 and satisfy x2 / X1 + X1 / x2 = 2, the value of M is 2. Let X1 and X2 be the two roots of the equation x ^ 2-x-1 = 0. （1）x1^2x2+x1x2^2 (2)(x1-x2)^2 (3)(x1+1/x2)(x2+1/x1)

1. If two of the univariate quadratic equations 2x ^ 2-mx + 4 = 0 are x1, X2 and satisfy x2 / X1 + X1 / x2 = 2, the value of M is 2. Let X1 and X2 be the two roots of the equation x ^ 2-x-1 = 0. （1）x1^2x2+x1x2^2 (2)(x1-x2)^2 (3)(x1+1/x2)(x2+1/x1)

1. From Veda's theorem, we get
x1+x2=-b/a=m/2
x1*x2=c/a=4/2=2
x2/x1+x1/x2
=（x1*x1+x2*x2）/x1*x2
=［（x1+x2）（x1+x2）-2x1*x2］/x1*x2
=（m/2*m/2-2*2）/2
=1/8*mm-2
=2
That is, mm = 32
∴m=±4√2
2,
（1）x1^2x2+x1x2^2
=x1*x2*（x1+x2）
=-1*1
=-1
(2) (x1-x2)^2
=（x1+x2）^2-4x1*x2
=1*1-4*（-1）
=5
(3) (x1+1/x2)(x2+1/x1)
=（x1*x2+1）（x1*x2+1）/（x1*x2）
=（-1+1）（-1+1）/1
=0
It's hard work,

We know that X1 + x2 = - B of a, X1 times x2 = C of a, ask x1-x2 =?
The above is the calculation. Everything is the law. What is the law of the problem?

(x1-x2) & sup2; = X1 & sup2; - 2x1x2 + x2 & sup2; = X1 & sup2; + 2x1x2 + x2 & sup2; - 4x1x2 = (x1 + x2) & sup2; - 4x1x2 = B & sup2; / A & sup2; - 4C / a = (B & sup2; - 4ac) / A & sup2; so x1-x2 = ± √ (B & sup2; - 4ac) / A

When m is a value, the equation (m-1) times the square of X - 2x + 3 = 0 has one root, and the other is a negative root. Which root has the largest absolute value?

It must be a positive root and a negative root. If both roots are less than 0, then multiplication = 3 / (m-1) should be greater than 0, so M-1 > 0. On the other hand, the sum of two roots = 2 / (m-1) should be less than 0 M-1