Mohon bantuannya yaaa (soal um)

[tex]\displaystyle x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2\\x_1^2+x_2^2=\left(-\frac ba\right)^2-\frac{2c}{a}\\x_1^2+x_2^2=\left(-\frac{a-2}1\right)^2-\frac{2(-a)}{1}\\x_1^2+x_2^2=a^2-4a+4+2a\\x_1^2+x_2^2=a^2-2a+4\\\text{karena minimum maka }\\a=-\frac{-2}{2(1)}\\a=1\\\\12(x_1+x_2-x_1x_2)=12\left(-\frac ba-\frac ca\right)\\12(x_1+x_2-x_1x_2)=12\left(-\frac{1-2}1-\frac{-1}1\right)\\12(x_1+x_2-x_1x_2)=12(1+1)\\12(x_1+x_2-x_1x_2)=24\\\\x_1^2+x_2^2=a^2-2a+4\\x_1^2+x_2^2=1^2-2(1)+4\\x_1^2+x_2^2=1-2+4\\x_1^2+x_2^2=3[/tex]

[tex]\displaystyle u_2=12(x_1+x_2-x_1x_2)\\ar=24\\\\u_5=x_1^2+x_2^2\\ar^4=3\\ar\cdot r^3=3\\24r^3=3\\r^3=\frac18\\r=\frac12\\\\ar=24\\a\cdot\frac12=24\\\boxed{\boxed{a=48}}[/tex]

x^2 + (a - 2)x - a = 0
x1^2 + x2^2 ==> minimum
= (x1 + x2)^2 - 2x1.x2
= (-(a - 2)/1)^2 - 2 (-a)/1
= a^2 - 4a + 4 + 2a
= a^2 - 2a + 4
Kita cari titik stasioner nya agar nilainya minimum dg turunan pertama
=> 2a - 2 = 0
=> 2a = 2
=> a = 1
x^2 + (a - 2)x - a = 0
x^2 + (1 - 2)x - 1 = 0
x^2 - x - 1 = 0
x1 + x2 = -(-1)/1 = 1
x1.x2 = c/a = -1/1 = -1
Barisan geometri
U2 = 12(x1 + x2 - x1.x2) = 12(1 - (-1)) = 12(2) = 24
U5 = x1^2 + x2^2 = (x1 + x2)^2 - 2x1.x2 = 1^2 - 2(-1) = 1 + 2 = 3

U5/U2 = 3/24
(ar^4)/(ar) = 1/8
r^3 = (1/2)^3
r = 1/2

U2 = 24
ar = 24
a(1/2) = 24
a = 48 => suku pertama