Given that x2 + 3x-1 = 0, find the value of x2 + 1 / x2 + 2x-2 / X

Given that x2 + 3x-1 = 0, find the value of x2 + 1 / x2 + 2x-2 / X

Hello, X & # 178; + 3x-1 = 0 divide both sides by the square of XX + 3-1 / x = 0x-1 / x = - 3 to get X & # 178; + 1 / X & # 178; - 2 = 9x & # 178; + 1 / X & # 178; = 11x & # 178; + 1 / X & # 178; + 2x-2 / x = x & # 178; + 1 / X & # 178; + 2 (x-1 / x) = 11 + 2 * (- 3) = 5

Given x2 + 3x + 1 = 0, find the value of X & # 178; + 1 / X & # 178

What a wonderful question! It almost baffled me!
∵x∧2＋3x＋1＝0
X (x + 3) = - 1
So x + 3 = - 1 / X
The two sides are squared,
（x＋3）∧2＝1/x∧2
So the formula
x∧2＋1/x∧2
＝x∧2＋（x＋3）∧2
＝x∧2＋x∧2＋6x＋9
＝2x∧2＋6x＋9
＝2（x∧2＋3x）＋9
＝2×（－1）＋9
＝7.

Known: (x-1) (x + 1) (X-2) (x-4) = (x2-3x) twice + a (x2-3x) + B, find the value of a, B

1. Special value substitution method
x=1
0=4-2a+b
x=3
-8=b
a=-2,b=-8
2. Common method:
(X-1)(X+1)(X-2)(X-4)
=(x-1)(x-2)(x+1)(x-4)
=(x^2-3x+2)(x^2-3x-4)
=(x^2-3x)^2-2(x^2-3x)-8
a=-2,b=-8