diketahui persamaan kuadrat 2x^2 + qx + q - 1 = 0 merupakan akar-akar x1 dan x2 jika x1^2 + x2^2 = 4 tentukan : a. nilai q b. nilai x1 + x2 dan x1.x2

[tex]2x^{2} + qx + q - 1 = 0 [/tex]
memiliki akar-akar [tex]x_{1}[/tex] dan [tex]x_{2}[/tex]
[tex]x_{1}^{2}+x_{2}^{2} = 4[/tex]

---- Identifikasi persamaan kuadrat ----
[tex]a = 2 \\ b = q \\ c = (q-1) \\[/tex]

---- Menentukan q ----
[tex]x_{1} + x_{2} = \frac{-b}{a} \\ x_{1} + x_{2} = \frac{-q}{2} \\ \\ x_{1}x_{2} = \frac{c}{a} \\ x_{1}x_{2} = \frac{(q-1)}{2}[/tex]

Diketahui [tex]x_{1}^{2}+x_{2}^{2} = 4[/tex]
[tex]x_{1}^{2}+x_{2}^{2} = 4\\ (x_{1}+x_{2})^{2}-2(x_{1}x_{2}) = 4 \\ (\frac{-q}{2})^{2} - 2(\frac{(q-1)}{2}) = 4 \\ \frac{q^{2}}{4} - (q - 1) = 4 \\ \frac{q^{2}}{4} - q + 1 = 4 \\ \frac{q^2-4q+4}{4} = 4 \\ q^{2}-4q+4 = 16 \\ q^{2} - 4q + 4 - 16 = 0 \\ q^{2} - 4q - 12 = 0 \\ (q + 2)(q - 6) = 0 \\ q = -2 | q = 6[/tex]

---- Menentukan [tex]x_{1} + x_{2}[/tex] dan [tex]x_{1}x_{2}[/tex] ----
Jika [tex]q = -2[/tex]
[tex]x_{1}+x_{2} = \frac{-q}{2} \\ x_{1}+x_{2} = \frac{-2}{2} \\ x_{1}+x_{2} = -1 \\ \\ x_{1}x_{2} = \frac{(q-1)}{2} \\ x_{1}x_{2} = \frac{-3}{2} [/tex]

Jika [tex]q = 6[/tex]

[tex]x_{1}+x_{2} = \frac{-q}{2} \\ x_{1}+x_{2} = \frac{-6}{2} \\ x_{1}+x_{2} = -3 \\ \\ x_{1}x_{2} = \frac{(q-1)}{2} \\ x_{1}x_{2} = \frac{5}{2} \\ [/tex]