How do you solve #4x² - 4x – 1 = 0#?

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Answer 1 · 2016-03-21T13:38:15

#x=(1+sqrt2)/(2)#

#color(blue)(4x^2-4x-1=0#

This is a Quadratic equation (in form #ax^2+bx+c=0#)

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

Where

#color(red)(a=4,b=-4,c=-1#

#rarrx=(-(-4)+-sqrt(-4^2-4(4)(-1)))/(2(4))#

#rarrx=(4+-sqrt(-4^2-4(4)(-1)))/(8)#

#rarrx=(4+-sqrt(16-(-16)))/(8)#

#rarrx=(4+-sqrt(16+16))/(8)#

#rarrx=(4+-sqrt(32))/(8)#

#rarrx=(4+-sqrt(16*2))/(8)#

#rarrx=(4+-4sqrt2)/(8)#

#rarrx=(cancel(4)^1+-cancel(4)^1sqrt2)/(cancel8)^2#

#color(green)(rArrx=(1+-sqrt2)/(2)#