Persamaan kuadrat x^2+ax+5a+3=0 mempunyai dua akar real alfa dan beta jika alfa+2beta=3 maka konstanta a=

[tex]\displaystyle x_1+2x_2=3\\x_1+x_2+x_2=3\\\frac{-a}{1}+x_2=3\\-a+x_2=3\\x_2=3+a\\\\x_1x_2=\frac{5a+3}{1}\\x_1(3+a)=5a+3\\x_1=\frac{5a+3}{a+3}\\\\x_1+x_2=\frac{-a}{1}\\\frac{5a+3}{a+3}+a+3=-a\\\frac{5a+3}{a+3}+2a+3=0\\\frac{5a+3+(a+3)(2a+3)}{a+3}=0\\5a+3+2a^2+9a+9=0\\2a^2+14a+12=0\\a^2+7a+6=0\\(a+1)(a+6)=0\\\boxed{\boxed{a=-1\vee a=-6}}[/tex]

x² + ax + (5a + 3) = 0

Berdasarkan Teoremq Vietta
X1 + X2 = -b/a
X1 + X2 = -a
A + β = -a
A = -β - a

X1. X2 = c/a
X1 . X2 = 5a + 3
A . β = 5a + 3

Maka...
A + 2β = 3
-β - a + 2β = 3
β = 3 + a

A = -β - a
A = -(3 + a) - a
A = -3 - 2a

Maka...
A . β = 5a + 3
(-3 - 2a)(3 + a) = 5a + 3
-9 - 3a - 6a - 2a² = 5a + 3
-2a² - 14a - 12 = 0
a² + 7a + 6 = 0
(a + 6)(a + 1) = 0
a = -6
a = -1