persamaan kuadrat x²+ ax + 5a + 3=0 mempunyai dua akar real α dan . Jika α + 2β = 3 maka konstanta a=

Materi: Persamaan Kuadrat

Pembahasan d gambar.

Perbaikan:
(-3-2a)(a+3)=5a+3
-2a^2-14a-12=0
a^2+7a+6=0
(a+6)(a+1)=0
a= -6 atau a=-1.

[tex]x^2 + ax + (5a + 3) = 0\\\\Dik :\\a = 1\\b =a\\c = 5a + 3\\\\Jawab :\\TeoremaVietta\\ \alpha . \beta = \frac{c}{a} = 5a + 3\\ \alpha + \beta = -\frac{b}{a} = -a\\\\ \alpha + 2 \beta = 3\\ (\alpha + \beta) + \beta = 3\\(-a) + \beta= 3\\ \beta = 3 + a\\\\ (\alpha + \beta) + \beta = 3\\( \alpha + 3 + a) + (3 + a) = 3\\ \alpha + 2a + 6 = 3\\ \alpha = -2a - 3\\\\ \alpha . \beta = 5a + 3\\(-2a - 3)(3 + a) = 5a + 3\\-6a - 2a^2 - 9 - 3a = 5a + 3\\2a^2 + 6a + 5a + 12 + 3a= 0\\2a^2 + 14a + 12= 0\\(2a + 2)(a + 6) = 0\\a =-1\\a= -6[/tex]

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