Akar-akar dari persamaan kuadrat x²+5x-6=0 adalah 1 dan -6

Cara:

1. Memfaktorkan

x² + 5x - 6 = 0

(x - 1) ( x + 6) = 0

x - 1 = 0 V x + 6 = 0

x1 = 1 V x2 = -6

2. Rumus abc

x1,x2

= -b ± √b²-4ac / 2a

= -5 ± √5² - 4(1)(-6) / 2(1)

= -5 ± √25 - (-24) / 2

= -5 ± √25 + 24 / 2

= -5 ± √49 / 2

= -5 ± 7 / 2

x1 = -5 + 7 / 2 = 2 / 2 = 1

x2 = -5 - 7 / 2 = -12/2 = -6

3. Kuadrat sempurna

x² + 5x - 6 = 0

x² + 5x + (5/2)² = 6 + (5/2)²

( x + 5/2)² = 6 + 25/4

( x + 5/2)² = 24/4 + 25/4

( x + 5/2)² = 49/4

x + 5/2 = √49/4

x + 5/2 = 7/2

x1 = 7/2 - 5/2 = 2/2 = 1

x2 = -7/2 - 5/2 = -12/2 = -6

Jadi, akar-akarnya adalah 1 dan -6