How do you solve #2x^2 + 8x + 7 = 0#?

1 Answer
Mar 26, 2016

Answer:

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

Explanation:

Completing the Square

Multiply by #2# to make the leading term into a perfect square, complete the square, then use the difference of square identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=(x+2)# and #b=sqrt(2)# as follows:

#0 = 4x^2+16x+14#

#=(2x+4)^2-16+14#

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

#=((2x+4)-sqrt(2))((2x+4)+sqrt(2))#

#=(2x+4-sqrt(2))(2x+4+sqrt(2))#

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

#=4(x+2-sqrt(2)/2)(x+2+sqrt(2)/2)#

Hence:

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

#color(white)()#
Quadratic Formula

#2x^2+8x+7# is in the form #ax^2+bx+c# with #a=2#, #b=8# and #c=7#.

This has zeros given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#=(-8+-sqrt(8^2-(4*2*7)))/(2*2)#

#=(-8+-sqrt(64-56))/4#

#=(-8+-sqrt(8))/4#

#=(-8+-sqrt(2^2*2))/4#

#=(-8+-2sqrt(2))/4#

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