Question

(x+1)(x+2)(x+3)(x+4)+1=0. , find x​

Answer 1

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Answer 2

Answer:

$(x + 1)(x + 2)(x + 3)(x + 4) + 1 = 0 \\ x(x + 2) + 1(x + 2) + x(x + 4) + 3(x + 4) + 1 = 0 \\ x {}^{2} + 2x + x + 2 + x {}^{2} + 4x + 3x + 12 + 1 = 0 \\ x {}^{2} + 3x + 2+ x {}^{2} + 7x + 12 + 1 = 0 \\2 x {}^{2} + 10x + 15 = 0 \\ comparing \: with \: ax { }^{2} + bx + c = 0 \: we \: get \\ a = 2 \\ b = 10 \\ c = 15 \\ using \: formula \: method \\ b {}^{2} - 4ac = 10 {}^{2} - 4(2)(15) \\ b {}^{2} \: { - 4ac} = 100 - 120 \\ b {}^{2} - 4ac = - 20 \\ therefore \: the \: roots \: would \: not \: be \: equal \\ since \: b { - 4ac}^{2} < 0$

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