How do you solve #(x - 2) ( x - 3) ( x + 4) ( x + 5) = 44#?

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Answer 1 · 2016-09-13T10:50:59

#x=-1-sqrt5# or #x=-1+sqrt5# or

#x=-1-2sqrt5# or #x=-1+2sqrt5#

Let us regroup #(x-2)(x-3)(x+4)(x+5)=44# as

#(x-2)((x+4)(x-3)(x+5)=44# or

#(x^2+2x-8)(x^2+2x-15)=44# ............(A)

Now let #x^2+2x=v#, then (A) becomes

#(v-8)(v-15)=44# or

#v^2-23v+120=44# or #v^2-23v+76=0# or

#v^2-19v-4v+76=0#

#v(v-19)-4(v-19)=0# or

#(v-4)(v-19)=0#

Hence #v=4# or #v=19#

But #v=x^2+2x# hence we have #x^2+2x=4# or #x^2+2x=19#

#x^2+2x-4=0# or #x^2+2x-19=0#

Using quadratic formula for #x^2+2x-4=0# gives #x=(-2+-sqrt(2^2-4xx1xx(-4)))/2=(-2+-sqrt20)/2=-1+-sqrt5#

and #x^2+2x-19=0# gives #x=(-2+-sqrt(2^2-4xx1xx(-19)))/2=(-2+-sqrt80)/2=-1+-2sqrt5#

Hence #x=-1-sqrt5# or #x=-1+sqrt5# or

#x=-1-2sqrt5# or #x=-1+2sqrt5#