Jika x1 dan x2 akar PK 4x²+bx+4=0, b≠0 maka x1^-1 + x2^-1=16(x1³+x2³) berlaku b-b² sama dengan..

# 4x² + bx + 4 = 0
# (1/x1) +(1/x2) = 16(x1^3 + x2^3)

b-b² = ?

jawaban:

x1.x2 = c/a = 4/4 = 1

(1/x1) + (1/x2) = 16(x1^3 + x2^3)
(x1+x2)/(x1.x2) =16((x1+x2)(x1²+x2²)-(x1.x2)(x1+x2)
(x1+x2)/(x1.x2) =16(((x1+x2))((x1²+x2²)-(x1.x2)))
>>>subtitusikan x1.x2 = 1 ...
x1+x2 = 16 (((x1+x2))((x1²+x2²)-1)))
>>>eliminasi x1+x2 ...
1 = 16 ((x1²+x2²)-1)
1 = 16 (x1²+x2²) - 16
17 = 16 ((x1+x2)²-2x1.x2)
17 = 16 (x1+x2)²-32
49 = 16 (x1+x2)²
49/16 = (x1+x2)²
x1+x2 = 7/4

x1+x2 = -b/a
7/4 = -b/4
-b = 7
b = -7

b - b² = -7 - (-7)²
= -7-49
=-56

maaf kalau salah

[tex]\displaystyle \frac{1}{x_1}+\frac{1}{x_2}=16(x_1^3+x_2^3)\\\frac{x_1+x_2}{x_1x_2}=16(x_1+x_2)(x_1^2-x_1x_2+x_2^2)\\\frac{1}{\frac ca}=16(x_1^2-\frac ca+x_2^2)\\\frac{1}{1}=16(x_1^2-1+x_2^2)\\1=16x_1^2-16+16x_2^2\\17=16x_1^2+16x_2^2\\17=16(x_1^2+x_2^2)\\17=16((x_1+x_2)^2-2x_1x_2)\\17=16\left(\left(-\frac ba\right)^2-2\frac ca\right)\\17=16\left(\frac {b^2}{16}-2\right)\\17=b^2-32\\49=b^2\\\pm7=b\\\\b-b^2=7-49\vee-7-49\\\boxed{\boxed{b-b^2=-42\vee-56}}[/tex]