Question

$(x + 1)(x + 2)(x + 3)(x + 4) = 120 \: \: prove \: \: that$
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Answer 1

Answer:

$(x + 1)(x + 2)(x + 3)(x + 4) = 120 \\ \\ = ((x + 1)(x + 4))(x + 2)(x + 3) = 120 \\ \\ = ( {x}^{2} + 5x + 4)( {x}^{2} + 5x + 6) = 120 \\ \\ (a + 4)(a + 6) - 120 = 0( {x}^{2} + 5x = a \: \: ok) \\ \\ {a}^{2} + 4a + 6a + 24 - 120 = 0 \\ \\ = {a}^{2} + 10a - 96 = 0 \\ \\ = {a}^{2} + 16a - 6a - 96 = 0 \\ \\ a(a + 16) - 6(a + 16) = 0 \\ \\ (a + 16)(a - 6) = 0 \\ a + 16 = 0 \: \: or \: \: a - 6 = 0 \\ \\ ifa + 16 = 0 \: \: then \: \: {x}^{2} + 5x + 16 = 0 \\ \\ {5}^{2} - 4 \times 1 \times 16 = 25 - 64 = - 39 < 0so \: \: the \: \: element \: \: are \: \: not \: \: real. \\ \\ if \: \: a - 6 = 0 \: \: then \: \: {x}^{2} + 5x - 6 = 0 \\ \\ x = \frac{ \sqrt[ - 5 + - ]{ {5}^{2} - 4 \times 1 \times ( - 6) } }{2 \times 1} \\ \\ = \frac{ - 5 + - \sqrt{ 25 + 24} }{2 } \\ \\ = \frac{ \sqrt[ - 5 + - ]{49} }{2} \\ \\ \frac{ - 5 + - 7}{2} \\ \\ x = \frac{ - 5}{7} or \: \: \frac{ - 5 - 7}{2} \\ \\ = x = 1or - 6$

Answer 2

Answer:

120=(x+1)(x+2)(x+3)(x+4)=((x+1)(x+4))((x+2)(x+3))

=((x2+5x+5)−1)((x2+5x+5)+1)…(1)

=(x2+5x+5)2−1,

we get x2+5x+5=±11.

Thus x2+5x−6=0 or x2+5x+16=0. The first of these give x=1 or x=−6. The second of these give x=12(−5±−39−−−−√).

It is easy to verify that x=1 and x=−6 are both solutions; in fact, 120=2⋅3⋅4⋅5=(−5)⋅(−4)⋅(−3)⋅(−2).

That the second pair of complex conjugates is also a solution can be best verified by replacing x2+5x by −16 in eqn. (1), say.

There are two real solutions, x=1 and x=−6, and two non-real solutions x=12(−5±−39−−−−√). ■