Question

Diketahui p dan q adalah akar-akar persamaan kuadrat 2x² - x + 4 = 0. Persamaan kuadrat yang akar akarnya
[tex] \frac{p}{q + 1} \: dan \: \frac{q}{p + 1} [/tex]
adalah...​

Answer 1

p + q = -(-1)/2 = 1/2

pq = 4/2 = 2

a + b = p/(q + 1) + q/(p + 1)

a + b = (p(p + 1) + q(q + 1))/(p + 1)(q + 1)

a + b = (p^2 + p + q^2 + q)/(pq + p + q + 1)

a + b = ((p + q)^2 - 2pq + p + q)/(2 + 1/2 + 1)

a + b = ((1/2)^2 - 4 + 1/2)/(3 + 1/2))

a + b = (1/4 + 1/2 - 4)/(3 1/2)

a + b = (3/4 - 4)/(3 1/2)

a + b = (3/4 - 16/4)/(7/2)

a + b = -13/4 x 2/7

a + b = -13/2 x 1/7

a + b = -13/14.

ab = (p/(q + 1))(q/(p + 1))

ab = pq/((q + 1)(p + 1))

ab = pq/(pq + p + q + 1)

ab = 2/(2 + 1/2 + 1)

ab = 2/(3 1/2)

ab = 2/(7/2)

ab = 4/7.

x^2 - (a + b)x + ab = 0

x^2 - (-13/14)x + 4/7 = 0

x^2 + 13/14x + 4/7 = 0

14x^2 + 13x + 8 = 0.

Opsi B.