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.

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