How do you solve #2x^2 + 5x -3=0# by completing the square?

1 Answer
George C.
Apr 25, 2016

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

Explanation:

We will use the difference of squares identity, which can be written:

#a^2-b^2=(a-b)(a+b)#

First, to reduce the amount of arithmetic involving fractions, first multiply through by #2^3 = 8# to get:

#0 = 16x^2+40x-24#

#= (4x)^2+2(5)(4x)-24#

#= (4x+5)^2-25-24#

#=(4x+5)^2-7^2#

#=((4x+5)-7)((4x+5)+7)#

#=(4x-2)(4x+12)#

#=(2(2x-1))(4(x+3))#

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

Hence:

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

#color(white)()#
Why did I premultiply by #8#?

One factor of #2# makes the leading term into a perfect square, then the additional factor #2^2 = 4# compensates for the middle term being an odd number.

Otherwise, we might proceed as follows:

#0 = 2x^2+5x-3#

#=2(x^2+5/2x-3/2)#

#=2((x+5/4)^2-25/16-3/2)#

#=2((x+5/4)^2-49/16)#

#=2((x+5/4)^2-7^2/4^2)#

#=2((x+5/4)-7/4)((x+5/4)+7/4)#

#=2(x-2/4)(x+12/4)#

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

Hence:

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

#color(white)()#
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