Jika 2p+1 dan q-3 adalah akar akar persamaan x2- (q-6)x+3p=0 maka nilai p+ q adalah

x1 = 2p + 1
x2 = q - 3
x1 + x2 = -b/a
2p + 1 + q - 3 = -(-(q - 6))/1
2p + q - 2 = q - 6
2p = q - q - 6 + 2
2p = -4
p = -2
x1 = 2(-2) + 1 = -3 substitusi ke persamaan
x² - (q - 6)x + 3p = 0
(-3)² - (q-6)(-3) + 3(-2) = 0
9 - (-3q + 18) - 6 = 0
9 + 3q - 18 - 6 = 0
3q - 15 = 0
3q = 15
q = 5
p + q = -2 + 5 = 3

X²-(q-6)x +3p=0
X1 = 2p +1
X2 =q-3
Ditanya nilai p+q

Diket
X²-(q-6)x +3p=0
a= 1
b= -(q-6)
= -q+6
c= 3p

X1+X2 = -b/a
2p +1 +q-3 = - (-q+6)/1
2p+q-2 = q-6
2p=-4
p=-2

X1 . X2 = c/a
(2p+1)(q-3)= 3p/1
(2.-2+1)(q-3) = 3.-2
-3q +9 = -6
-3q = -15
q= 5

p+q = -2+5 = 3