Interesting Equation

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Find half the sum of solutions to the equation: ( x + 1 ) ( x + 2 ) ( x + 3 ) ( x + 4 ) = − 1 (x+1)(x+2)(x+3)(x+4) = -1

-10 25 0 -5

Jam M by Jam M , 39, USA

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1 solution

Jam M
Feb 29, 2020

( x + 1 ) ( x + 2 ) ( x + 3 ) ( x + 4 ) = − 1 ( x + 1 ) ( x + 4 ) ( x + 2 ) ( x + 3 ) + 1 = 0 ( x 2 + 5 x + 4 ) ( x 2 + 5 x + 6 ) + 1 = 0 (x+1)(x+2)(x+3)(x+4) = -1 \\ (x+1)(x+4)(x+2)(x+3) + 1 = 0 \\ (x^2 + 5x + 4)(x^2 + 5x + 6) + 1 = 0 Let y = x 2 + 5 x y = x^2 + 5x so that ( y + 4 ) ( y + 6 ) + 1 = 0 y 2 + 10 y + 25 = 0 ( y + 5 ) 2 = 0 y = − 5 double root (y+4)(y+6) + 1 = 0 \\ y^2 + 10y + 25 = 0 \\ (y+5)^2 = 0 \\ y = -5 \; \mbox{double root} x 2 + 5 x = − 5 x 2 + 5 x + 5 = 0 x = − 5 ± 5 2 each a double root x^2 + 5x = -5 \\ x^2 + 5x + 5 = 0 \\ x = \dfrac{-5 \pm \sqrt{5}}{2} \; \mbox{each a double root} The solutions to the given equation have sum − 10 -10 . Half the sum is − 5 -5 .

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