Quadratic Root
For a and b are two prime numbers x 2 − a x + b = 0 have two different positive integer roots. What is a + 2 b .
The answer is 7.
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2 solutions
x 2 − a x + b = 0 x 1 , x 2 is two different roots. x 1 ⋅ x 2 = b but b is a prime. So either x 1 or x 2 is 1 and the other is b . We take x 1 = b and x 2 = 1 x 1 + x 2 = a b + 1 = a a − b = 1 a = 3 ; b = 2 a + 2 b = 7
The wording of the problem says the quadratic should have "two different real roots". You need to specify that you are looking for integer roots, otherwise any pair of primes satisfying a 2 > 4 b is a solution (for example, a = 5 , b = 2 gives distinct real roots).
From the given conditions of the problem, a^2>4b. There are infinitely many solutions to this inequality with a, b prime. For the roots to be positive integers, a=3 and b=2 is the only solution. (Since a^2-4b=c^2 with c a positive integer implies (a+c)(a-c)=4b. Since b is prime, either a+c or a-c must be 4. In either case, a=(b+4)/2, which holds for a=3 and b=2 only)