How do you solve by completing the square for #x^2+ 3/2x = 3#?

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Answer 1 · 2018-03-20T08:21:59

#x=+sqrt(57/16)-3/4 = 1.1374#

or

#x=-sqrt(57/16)-3/4 = -2.6374#

#x^2+3/2x=3#

#=> x^2+3/2x-3=0#

#=> (x+3/4)^2-(3/4)^2-3=0#

#=> (x+3/4)^2-(9/16)-3=0#

#=> (x+3/4)^2-(9/16)-(48/16)=0#

#=> (x+3/4)^2-(57/16)=0#

#=> (x+3/4)^2=57/16#

#=> x+3/4=+-sqrt(57/16)#

#=> x=+-sqrt(57/16)-3/4#

Therefore, x is either

#x=+sqrt(57/16)-3/4 = 1.1374#

#x=-sqrt(57/16)-3/4 = -2.6374#

Answer 2 · 2018-03-20T08:27:47

#x=(-3+-sqrt57)/4#

#x^2+3/2 x =3#

Add #(3/4)^2# both sides.

#=> x^2+3/2x+(3/4)^2 = 3+(3/4)^2#

#=> (x)^2 + 2xx x xx3/4 + (3/4)^2 = 3+9/16#

#=> (x+3/4)^2 = 57/16#

#=> x+3/4=(+-sqrt57)/4#

#=> x=(+-sqrt57)/4 - 3/4#

#=> x=(-3+-sqrt57)/4#