How do you solve #x^2-2x-4=0# by factoring?

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Answer 1 · 2016-10-26T12:51:35

#x=1+sqrt5#
OR
#x=1-sqrt5#

The equation can be easily solved by factoring #x^2-2x-4# which is done by applying the quadratic formula.

#color(blue)(delta=b^2-4ac)=(-2)^2-4(1)(-4)=4+16=20#

#color(green)(x_1=(-b+sqrtdelta)/(2a))=(2+sqrt20)/2=(2+sqrt(4*5))/2=(2+2sqrt5)/2#
#x_1=(1+sqrt5)#

#color(brown)(x_2=(-b-sqrtdelta)/(2a))=(2-sqrt20)/2=(2-sqrt(4*5))/2=(2-2sqrt5)/2#
#x_2=(1-sqrt5)#

#x^2-2x-4=0#
#rArr(x-color(green)(x_1))(x-color(brown)(x_2))=0#

#rArr(x-(1+sqrt5))(x-(1-sqrt5))=0#

#rArr#
#x-(1+sqrt5)=0rArrx=1+sqrt5#
OR
#x-(1-sqrt5)=0rArrx=1-sqrt5#