f(x) = x³-4x²+5x+6 (x1)²+(x2)²+(x3)² = ...

Semoga bermanfaat ..

Itu kalo saya gak salah hitung, kalo salah hapus saja ..

f(x) = x³-4x²+5x+6 => a=1 , b=-4 , c=5 , d=6
-pertama kita misal-kan X1 = -X2
X1+X2+X3 = -b/a
X1+X2+X3 = -(-4)/1 = 4/1 = 4
-X2 + X2 +X3 =4 ==>X3 = 4 ,dan
X1.X2 + X2.X3 + X1.X3 = c/a
X1.X2 + X3 (X2+X1) = 5/1
X1.X2 + 4 (X2+(-X2)) = 5
X1.X2 + 4 = 5
X1.X2 = 1
(-X2).X2 = 1
-X2² = 1 -> X2 = √-1 dan,
X1.X2 = 1
X1.√-1 = 1 -> X1 = 1/√-1.
_setelah itu,substitusikan;
X1²+X2²+X3² = (1/√-1)² + (√-1)² + 4²
= -1 + (-1) + 16
= 14, alhasil 14.