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Soal nomor 3
Diketahui persamaan kuadrat 3x² − 12x + 2 = 0 adalah α dan β
Ditanya persamaan kuadrat baru yang akar-akarnya (α + 2) dan (β + 2)
Jawab: Jika 3x² − 12x + 2 = 0 maka a = 3, b = -12, c = 2
jadi α + β = - b/a = - -12/3 = 4
α . β = c/a = 2/3
Bentuk akar barunya adalah (α + 2) dan (β + 2)
x² - (x1 +x2)x + (x1 .x2) = 0
x² - (α + β + 4)x + (αβ + 2(α + β) + 4) = 0
x² - (4 + 4)x + (2/3 + 2(4) + 4) = 0
x² - 8x + 38/3 = 0
3x² - 24x + 38 = 0
Soal nomor 4
a. x² - 1 = 0
x² - 1 = (x + 1) (x - 1)
Maka x = 1 atau x = -1.
b. 4x² + 4x + 1 = 0
4x² + 4x + 1 = 0
(2x + 1 ) (2x + 1) = 0
Maka x = -1/2
c. -3x - 5x +2 = 0
3x + 5x - 2 = 0
(3x - 1) (x + 2) = 0
Maka x = 3x - 1 atau x = ⅓
x = x + 2 atau x = -2
d. 2x² - x - 3 = 0
(2x - 3) (x + 1) = 0
2x - 3 = 0 atau x + 1 = 0
Maka x = 3/2 atau x = -1
e. x² - x + ¼ = 0
(x - ½) (x - ½) = 0
Maka x = ½
Soal nomor 5
a. a. 3x² – 12 = 0
3x² – 12 = 0
Maka a = 3, b = 0, c = -12
Jadi Diskiriminan = 0² – 4(3)(–12) = 144
b. x² + 7x + 6 = 0
x² + 7x + 6 = 0
Maka a = 1, b = 7, c = 6
Jadi Diskiriminan = 7² – 4(1)(6) = 49 – 24 = 25
c. -3x² - 5x + 2 = 0
-3x² - 5x + 2 = 0
Maka a = -3, b = -5, c = 2
Jadi Diskiriminan = (–5)² – 4(–3)(2) = 25 + 24 = 49
Soal nomor 6
3x² – 5x + c = 0
Maka a = 3, b = -5
Jika D = b² - 4ac, D = 49
49 = (-5)² - 4.3.c
49 - 25 = -12.c
-12.c = 24
c = - 24/12
c = - 2
Soal nomor 7
3x2 = 2x – 4
Maka 3x2 – 2x + 4 = 0
