If p and q are primes and $x^{2}$ -px+q=0 has distinct positive integral roots, find p and q
0 bình luận về “If p and q are primes and $x^{2}$ -px+q=0 has distinct positive integral roots, find p and q”
Đáp án: p=3, q=2
Giải thích các bước giải:
Consider the equation: x² – px + q = 0
Because Pt has 2 distinct positive integers x1; x2 should follow Viète’s theorem with two 2 satisfied solutions
{x1 + x2 = p (1)
{x1.x2 = q = 1.q (2) (since q is prime)
From (2) we can assume that x1 = 1 <x2 = q
Substitute (1) 1 + q = p
Do p; q prime should have 2 cases:
– if q odd elements => p = q + 1 even elements => p = 2 => q = 1 instead of (1) has the equation: x² – 2x + 1 = 0. This equation has a dual solution x = 1 not satisfying
– if q is even, prime q = 2 => p = q + 1 odd prime => p = 3 instead of (1) has the equation: x² – 3x + 2 = 0. This equation has 2 solutions of pb: x1 = 1; x2 = 2 satisfies the problem
Đáp án: p=3, q=2
Giải thích các bước giải:
Consider the equation: x² – px + q = 0
Because Pt has 2 distinct positive integers x1; x2 should follow Viète’s theorem with two 2 satisfied solutions
{x1 + x2 = p (1)
{x1.x2 = q = 1.q (2) (since q is prime)
From (2) we can assume that x1 = 1 <x2 = q
Substitute (1) 1 + q = p
Do p; q prime should have 2 cases:
– if q odd elements => p = q + 1 even elements => p = 2 => q = 1 instead of (1) has the equation: x² – 2x + 1 = 0. This equation has a dual solution x = 1 not satisfying
– if q is even, prime q = 2 => p = q + 1 odd prime => p = 3 instead of (1) has the equation: x² – 3x + 2 = 0. This equation has 2 solutions of pb: x1 = 1; x2 = 2 satisfies the problem