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)()#
Which method do you prefer?

Related questions
See all questions in Completing the Square
Impact of this question
11242 views around the world
You can reuse this answer
Creative Commons License