How do you solve using the completing the square method #x^2 = (3/4)x - (1/8)#?

1 Answer
George C.
Mar 19, 2016

Multiply by #64# first to cut down on the fractions, then complete the square and use the difference of squares identity to find:

#x = 1/2# or #x=1/4#

Explanation:

The difference of squares identity can be written:

#A^2-B^2 = (A-B)(A+B)#

I use this below, with #A=(8x-3)# and #B=1#.

#color(white)()#
To match #ax^2+bx# with a square, you normally look at:

#a(x+b/(2a))^2 = ax^2+bx+b^2/(4a)#

In our example, we can rearrange the original equation to get one involving #a=1# and #b=-3/4#, which would lead us to looking at:

#(x-3/8)^2 = x^2-(3/4)x+9/64#

These fractions get a little painful, so let us multiply the original equation by #64# before we start and cut down on the fractions...

#x^2=(3/4)x-(1/8)#

becomes:

#64x^2=48x-8#

which we can rearrange as:

#0 = 64x^2-48x+8#

#=(8x)^2-2(8x)(3)+8#

#= (8x-3)^2-3^2+8#

#= (8x-3)^2-9+8#

#= (8x-3)^2-1#

#= (8x-3)^2-1^2#

#= ((8x-3)-1)((8x-3)+1)#

#= (8x-4)(8x-2)#

#= (4(2x-1))(2(4x-1))#

#= 8(2x-1)(4x-1)#

So #x=1/2# or #x=1/4#

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