Respuesta :
Hello,
3x²-x+9=0
==>x²-x/3+3=0
p+q=1/3
pq=3
p/q+q/p=[(p+q)²-2pq]/(pq)=[(1/3)²-2*3]/3=[1/9-6]/3=-53/27
3x²-x+9=0
==>x²-x/3+3=0
p+q=1/3
pq=3
p/q+q/p=[(p+q)²-2pq]/(pq)=[(1/3)²-2*3]/3=[1/9-6]/3=-53/27
3x²- x + 9 = 0
also x² - (1/3)x + 3 = 0
p,q are roots so
(x - p)(x - q) = x² - (1/3)x + 3 = 0
x² - (p+q)x + pq = x² - (1/3)x + 3 = 0
compare coefficients we have
p+q = 1/3 ... (1)
pq = 3 ... (2)
(1)/(2) ... (p+q)/pq = (1/3)/3
p/q + q/p = 1/9
also x² - (1/3)x + 3 = 0
p,q are roots so
(x - p)(x - q) = x² - (1/3)x + 3 = 0
x² - (p+q)x + pq = x² - (1/3)x + 3 = 0
compare coefficients we have
p+q = 1/3 ... (1)
pq = 3 ... (2)
(1)/(2) ... (p+q)/pq = (1/3)/3
p/q + q/p = 1/9