How do you solve by completing the square for #2x^2+10x-4=0#?

2 Answers
Mar 13, 2018

#2(x + 5/2)^2 - 33/2#

Explanation:

Take 2 out of the equation:
#2(x^2 + 5x - 2) = 0#

Complete the square in the brackets:
#2((x + 5/2)^2 - 25/4 - 2)#

Simplify the equation:
#2((x + 5/2)^2 - 33/4)#
#2(x + 5/2)^2 - 33/2#

Mar 13, 2018

#2x^2 +10x -4=0#

First divide by #2# so that we have #1x^2" "a=1#

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

#x^2 +5x " "=2" "larr# move the constant to the RHS

#x^2+5x+ (5/2)^2 = 2+(5/2)^2" "larr# add #(b/2)^2# to both sides.

By this process you have written the left side as a 'perfect square'
This step is the 'completing the square '- add in the missing term to create a square.

Write the left side as the square of a binomial.

#(x+5/2)^2 = 4 1/2#

#x +5/2 = +-sqrt(9/2)" "larr# find the square root of both sides.

Find the two possible solutions:

#x = 3/(+sqrt2) -2.5 = -0.379# (3 dp)

#x = 3/(-sqrt2)-2.5 = -4.624#