There are () rational numbers x that make (x ^ 2 + 3x) ^ 2-2 (x ^ 2 + 3x) - 8 equal to zero A.2 B.3 C.4 D.5

There are () rational numbers x that make (x ^ 2 + 3x) ^ 2-2 (x ^ 2 + 3x) - 8 equal to zero A.2 B.3 C.4 D.5

That is, (X & # 178; + 3x-4) (X & # 178; + 3x + 2) = 0
(x+4)(x-1)(x+1)(x+2)=0
So it's four
Choose C

If x2-3x + 1 = 0, then the value of x2 + 1x2 is ()
A. 8B. 7C. 3±52D. 7±52

From x2-3x + 1 = 0, we can get x2 + 1 = 3x. From the problem, X is not equal to 0, both sides divide by X, we can get: x2 + 1x = 3 ① It is also known that x 2 + 1 x 2 = x 2 + 2x · 1 x + (1 x) 2-2x · 1 x = (x + 1 x) 2-2 = (x 2 + 1 x) 2-2 ② Substituting ① into ②, the original formula is 32-2 = 7

Now a kind of operation is defined: a * b = AB + A-B, where a and B are rational numbers, and the value of [3x (- 4)] X2 is obtained

【3*(-4)】*2
=【3x(-4)+3-(-4)】*2
=(-5)*2
=(-5)x2+(-5)-2
=-17