Contoh
Akar-akar persamaan kuadrat x2 + (a – 1)x + 2 = 0 adalah α dan β. Jika α = 2β dan a > 0 maka nilai a = …
A. 2
B. 3
C. 4
D. 6
E. 8
Alternatif
Pembahasan:
α = 2β
(i) α ⋅ β = c/a
2β ⋅ β = 2/1
2β2 = 2
β2 = 1
β = ± 1
β = 1 atau β = –1
(ii) α + β = –
b/a
2β + β =
3β = 1 – a
3(–1) = 1 – a
a = 1 + 3
= 4
Jawaban : C
Contoh
Persamaan kuadrat 2x2 + 3x – 5 =
0, mempunyai akar-akar x1 dan x2. Persamaan
kuadrat baru yang akar-akarnya (2x1 –
3) dan (2x2 – 3) adalah …
A. 2x2 + 9x + 8 = 0
B. x2 + 9x + 8
= 0
C. x2 – 9x – 8
= 0
D. 2x2 – 9x + 8 = 0
E. x2 + 9x – 8
= 0
Persamaan kuadrat lama :
2x2 + 3x – 5 = 0, a = 2, b
= 3, c = – 5
Akar-akar persamaan kuadrat baru
(i)
α + β = 2 x1
– 3 + 2x2 – 3 =
= 2(x1 + x2)
– 6
= 2 (– b/a
) – 6
= 2 (−3/2)
– 6
= – 3 –
6
= – 9
(ii)
α · β = (2x1
– 3) (2x2 – 3)
= 4(x1
· x2) – 6x1 – 6x2 +
9
= 4(x1
· x2) – 6(x1 + x2)
+ 9
= 4 (c/a)
– 6(−b/a) + 9
= 4 (−5/2)
– 6 (−3/2) + 9
= – 10 +
9 + 9
= 8
Jadi, persamaan kuadrat barunya adalah:
x2 – (α + β)x
+ (α · β) = 0
⇔ x2
– (– 9)x + 8 = 0
⇔ x2
+ 9x + 8 = 0
Jawaban : B
Sumber
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