Rešite (x + 1) (x + 3) (x + 4) (x + 6) = 112?

Rešite (x + 1) (x + 3) (x + 4) (x + 6) = 112?
Anonim

Odgovor:

# x = -7 / 2 + -isqrt31 / 2 # ali # x = -7 / 2 + -sqrt57 / 2 #

Pojasnilo:

Naj združimo LHS as

# (x + 1) (x + 6) (x + 3) (x + 4) = 112

# => (x ^ 2 + 7x + 6) (x ^ 2 + 7x + 12) = 112 #

Zdaj pa pustite # u = x ^ 2 + 7x # in nato nad enačbo postane

# (u + 6) (u + 12) = 112 #

ali # u ^ 2 + 18u + 72 = 112 #

ali # u ^ 2 + 18u-40 = 0 #

ali # (u + 20) (u-2) = 0 # t.j. # u = 2 # ali #-20#

Tudi to tudi # x ^ 2 + 7x + 20 = 0 # t.j. #x = (- 7 + -sqrt (7 ^ 2-80)) / 2 # t.j. # x = -7 / 2 + -isqrt31 / 2 #

ali # x ^ 2 + 7x-2 = 0 # t.j. #x = (- 7 + -sqrt (7 ^ 2 + 8)) / 2 # t.j. # x = -7 / 2 + -sqrt57 / 2 #