The importance of discriminant (2)

Algebra Level 4

a ⋄ b = a 2 + 4 a b + b 2 − 2 a − 2 b + 9 a b + 6 \large a \diamond b = \frac{\sqrt{a^2+4ab+b^2-2a-2b+9}}{ab+6}

If ( ( ⋯ ( 2017 ⋄ 2016 ) ⋄ 2015 ) ⋄ ⋯ ⋄ 2 ) ⋄ 1 ) = c d (( \cdots (2017 \diamond 2016) \diamond 2015) \diamond \cdots \diamond 2) \diamond 1) = \dfrac{\sqrt{c}}{d} , where c c and d d are square-free positive integers , find c + d c+d .


The answer is 50.

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Tommy Li by Tommy Li , 22, Hong Kong

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1 solution

Tommy Li
Aug 6, 2017

Δ \Delta of a 2 + ( 4 b − 2 ) a + ( b 2 − 2 b + 9 ) = ( 4 b − 2 ) 2 − 4 ( b 2 − 2 b + 9 ) = 0 a^2+(4b-2)a+(b^2-2b+9) = (4b-2)^2-4(b^2-2b+9) = 0

b = 2 b = 2 or b = − 4 3 b = -\frac{4}{3} (rej.)

a ⋄ 2 = a 2 + 4 a ( 2 ) + 2 2 − 2 a − 2 ( 2 ) + 9 2 a + 6 = ( a + 3 ) 2 2 ( a + 3 ) = 1 2 a \diamond 2 = \dfrac{\sqrt{a^2+4a(2)+2^2-2a-2(2)+9}}{2a+6} = \dfrac{\sqrt{(a+3)^2}}{2(a+3)} = \dfrac{1}{2}

( ( ⋯ ( 2017 ⋄ 2016 ) ⋄ 2015 ) ⋄ ⋯ ⋄ 2 ) ⋄ 1 ) = 1 2 ⋄ 1 = 37 13 (( \cdots (2017 \diamond 2016) \diamond 2015) \diamond \cdots \diamond 2) \diamond 1) = \dfrac{1}{2} \diamond 1 = \dfrac{\sqrt{37}}{13}

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