How do you solve using the completing the square method #2x^2 + 8x + 1 = 0#?

2 Answers
P dilip_k
Feb 27, 2016

#2x^2+8x+1=0#
Dividing both sides by 2 we get
#=>x^2+4x+1/2=0#
#=>x^2+2*x*2+2^2-2^2+1/2=0#
#=>x^2+2*x*2+2^2=4-1/2=7/2#
#=>(x+2)^2=7/2#
#=>x+2=+-sqrt(7/2)#
#=>x=-2+-sqrt(7/2)#

Noah G
Feb 27, 2016

See solution below.

Explanation:

#2(x^2 + 4x + m) = -1#

#m = (b/2)^2#

#m = (4/2)^2#

#m = 4#

#2(x^2 + 4x + 4 - 4) = -1#

#2(x + 2)^2 - 4(2) = -1#

#2(x + 2)^2 = -1 + 8#

#(x + 2)^2 = 7/2#

#x + 2 = +-sqrt(7/2)#

#x = +-sqrt(7/2) - 2#

Don't forget the #+-# sign with the square root. Remember: a quadratic equation must always have two solutions.

Related questions
See all questions in Completing the Square
Impact of this question
2264 views around the world
You can reuse this answer
Creative Commons License