There are positive and such that the polynomial has two real roots which differ by 30. Find the least possible value of given and are positive integers.
This problem is not original.
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Rewrite the above quadratic equation as x 2 + 2 b x + 2 c = ( x + r ) ( x + ( r + 3 0 ) ) = 0 . This yields: b = 4 r + 6 0 , c = 2 r 2 + 6 0 r . Given that b , c ∈ N , we require:
b = 4 r + 6 0 > 0 ⇒ r > − 1 5 AND c = 2 r 2 + 6 0 r > 0 ⇒ r > 0 , r < − 3 0
of which r > 0 satisfies both conditions. Since r = 1 is the smallest positive integer in this instance, this yields b = 6 4 , c = 6 2 ⇒ b + c = 1 2 6 .