How do you solve #2x^2 + 8x + 7 = 0#?
1 Answer
Mar 26, 2016
#x = -2+-sqrt(2)/2#
Explanation:
Multiply by
#a^2-b^2 = (a-b)(a+b)#
with
#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#
Quadratic Formula
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#