How do you solve #(x+1)^2-9/4=0#?

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Answer 1 · 2017-02-06T21:32:05

#x=-5/2color(white)("XX")orcolor(white)("XX")x=+1/2#

#(x+1)^2-9/4=0#

#rarr (x+1)^2 = 9/4#

#rarr x+1 = +-3/2#

#rarr x=-1+-3/2#

#rarr x=-5/2color(white)("XX")orcolor(white)("XX")x=+1/2#

Answer 2 · 2017-02-06T21:34:10

Being a second-order equation, their are two solutions

#x=1/2# and #x=-5/2#

First, move the right-most term to the right side of the equation

#(x+1)^2=9/4#

Now, take the square root of each side. Don't forget that there will be two roots to the right side:

#x+1=+-3/2#

So, the answers are

#x=+3/2-1# or #x=1/2#
and

#x=-3/2-1# or #x=-5/2#

Answer 3 · 2017-02-06T21:39:52

#x = -2 1/2 " or " x = 1/2#

While you could multiply out the bracket and then simplify and factorise the quadratic trinomial, an easier method is to realize that

#(x+1)^2-9/4# can be factorised as the difference of squares.

#x^2 -y^2 = (x+y)(x-y)#

So, if #(x+1)^2-9/4 = 0#

then #(x+1+3/2)(x+1-3/2)=0#

Consider each factor being equal to 0.

#x+1+3/2 = 0" "rarr x = -5/2 = -2 1/2#

#x+1 -3/2 =0" "rarr x = 1/2#