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

1 Answer
Dec 4, 2016

Answer:

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

Explanation:

The difference of squares identity can be written:

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

We use this later with #a = (4x+3)# and #b = 7#

#color(white)()#
Given:

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

To delay having to work with fractions, first multiply this by #8#.

Why #8#?

We want to multiply by #2# to make the leading term into a perfect square, namely #4x^2 = (2x)^2#. Then the middle term becomes #6x=3(2x)#. We want the coefficient #3# of #(2x)# to be even too, so multiply by another factor of #2^2 = 4# to keep the leading term a perfect square.

Then we find:

#0 = 8(2x^2+3x-5)#

#color(white)(0) = 16x^2+24x-40#

#color(white)(0) = (4x)^2+2(3)(4x)+9-49#

#color(white)(0) = (4x+3)^2-7^2#

#color(white)(0) = ((4x+3)-7)((4x+3)+7)#

#color(white)(0) = (4x-4)(4x+10)#

#color(white)(0) = 4(x-1)2(2x+5)#

Hence:

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