How do you solve #4x²-1=0#?

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Answer 1 · 2016-09-03T11:10:22

1 or -1

#(2x-1)(2x+1)=0#
#2x-1=0,2x+1=0#
#2x=0+1,2x=0-1#
#2x=1,2x=-1#
#x=1/2,x=-1/2#
#S.S{+1/2,-1/2}#

Answer 2 · 2016-09-03T11:27:35

#x=+-1/2#

#color(green)("You only need to solve as a quadratic if you have a "bx)##color(green)("term in the standard form of "y=ax^2+bx+c)#

Given:#" "4x^2-1=0#

Add 1 to both sides

#4x^2=1#

Divide both sides by 4

#x^2=1/4#

Square root both sides

#x=+-sqrt(1/4) = +-1/2#