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

2 Answers
Jun 6, 2018

Answer:

#x_1=3#, #x_2=-1/2#

Explanation:

Completing the square means that you want to write #2x^2-5x-3=0#
on the form #(x-a)^2=b#

First, let's get rid of the constant in front of the 2nd degree term:
#x^2-5/2x-3/2=0#
We want this on the form
#(x-a)^2=b#
i.e. #x^2-2ax+a^2=b#
Comparing term for term, we see that
#2a=5/2#, i.e. #a=5/4#
#(x-5/4)^2=x^2-5/2x+25/16#
We, therefore, have
#(x^2-5/2x+25/16)-25/16-3/2=0#
or #(x^2-5/2x+25/16)=25/16+24/16=49/16#
#(x-5/4)^2=(7/4)^2#
#x-5/4=+-7/4#
#x=5/4+-7/4#
Therefore #x_1= 5/4+7/4=3#
#x_2=5/4-7/4=-1/2#

Answer:

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

Explanation:

Given:

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

On dividing this equatin by #2#, we get:

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

#x^2 -5/2x = 3/2#

Add #(5/4)^2# on both sides

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

[because #a^2 + b^2 + 2ab = (a+b)^2# ]

So

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

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

Taking the square root on both sides and solving it.

#x-5/4 = 7/4" "# or #" "x-5/4 = -7/4#

#x = 7/4+5/4" "# or #" "x = -7/4+5/4#

#x = 12/4" "# or #" "x = -2/4#

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