Jika selisih akar akar x^+2mx+(19+m)=0 adalah 2, maka nilai m^-m+3 = ...

Jawab :

Jawab bareng Kakakku To9ha :D

x² + 2mx + (19 + m) = 0
a = 1
b = 2m
c = 19 + m

D = b² - 4ac = 4m² - 4(19 + m)

x1 - x2 = 2
√D /a = 2
Kuadratkan kedua ruas
D/a² = 2²
4m² - 4(19 + m) = 4
4m² - 4m - 76 - 4 = 0
m² - m - 20 = 0
m² - m = 20

Maka,
m² - m + 3
= 20 + 3
= 23 ✔

jawab

x² + 2m x + (19+m) = 0
a= 1
b = 2m
c = 19 +m

x1 + x2 = -b/a= 2m
x1x2 = c/a = 19+m

x1- x2 = 2
(x1 - x2)² = 2²
x1² + x2² - 2x1 x2 = 4
(x1 + x2)² - 4 x1x2 = 4
(2m)² - 4(19+m) = 4
4m² - 76 - 4m  - 4= 0
4m² - 4m - 80 =0
m² - m - 20 = 0
(m - 5)(m +4)= 0
m = 5  atau m = - 4

nilai  P= m² - m + 3 
m= 5 → P = 25-5+ 3 = 23
m= - 4 → P = 16-(-4)+3 = 23