Answer:
x² -3/4x +1/4 = 0
Step-by-step explanation:
Consider the two equations in factored and expanded forms:
(x -p²)(x -q²) = x² -(p²+q²)x +p²q² = 0 ⇒ p²+q² = 1, p²q² = 16
and
(x -1/p)(x -1/q) = x² -(1/p+1/q)x +1/(pq) = 0
Consider the squares of the sum and product of roots:
constant term: (1/(pq))² = 1/(p²q²) = 1/16 ⇒ 1/(pq) = √(1/16) = 1/4
x-term: (1/p +1/q)² = (p +q)²/(pq)² = (p² +q² +2pq)/(p²q²)
= (p² +q²)/(p²q²) +2/(pq)
= 1/16 +2/√16 = 9/16 ⇒ (1/p +1/q) = √(9/16) = 3/4
Then the equation with roots 1/p and 1/q is ...
x² -3/4x +1/4 = 0
