How do you solve # 2x^2-3x-14=0# by completing the square?

1 Answer
Dec 22, 2016

Answer:

#x=7/2" "# or #" "x=-2#

Explanation:

The difference of squares identity can be written:

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

We will use this with #A=(4x-3)# and #B=11# later.

#color(white)()#
Given:

#2x^2-3x-14 = 0#

To avoid fractions as much as possible, let us premultiply this quadratic by #8 = 2*2^2#. One factor of #2# makes the leading term a perfect square, then the other #2^2# deals with the #2# denominator in "#b/(2a)#" which would otherwise cause us to have to work with #1/2#'s.

So:

#0 = 8(2x^2-3x-14)#

#color(white)(0) = 16x^2-24x-112#

#color(white)(0) = (4x)^2-2(4x)(3)+9-121#

#color(white)(0) = (4x-3)^2-11^2#

#color(white)(0) = ((4x-3)-11)((4x-3)+11)#

#color(white)(0) = (4x-14)(4x+8)#

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

#color(white)(0) = 8(2x-7)(x+2)#

Hence:

#x=7/2" "# or #" "x=-2#