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

Question

2122 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-11T08:55:09

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

Let us group LHS as

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

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

Now let #u=x^2+7x# and then above equation becomes

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

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

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

or #(u+20)(u-2)=0# i.e. #u=2# or #-20#

As such either #x^2+7x+20=0# i.e. #x=(-7+-sqrt(7^2-80))/2# i.e. #x=-7/2+-isqrt31/2#

or #x^2+7x-2=0# i.e. #x=(-7+-sqrt(7^2+8))/2# i.e. #x=-7/2+-sqrt57/2#