How do you solve #x ^ { 2} - 4x - 12= 0#?

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Answer 1 · 2017-11-18T16:41:25

the given equation is #x^2-4x-12=0rArrx^2-6x+2x-12=0rArrx(x-6)+2(x-6)=0rArr(x-6)(x+2)=0rArrx=-2,6#

Answer 2 · 2017-11-18T16:42:54

#x=6 and x=-2" are solutions"#

Observe that
#2xx6=12; (+2)xx(-6)=-12 " and that" +2-6=-4#

#color(green)(x^2-4x-12=0color(white)("d")->color(white)("d")(x+2)(x-6)=0)#

For this to work we have 2conditions.

#color(green)( color(white)("dd")0color(white)("dd")xx(x-6)=0 ->(x+2)color(white)("d")=0" thus "x=-2)#

#color(green)((x+2)xxcolor(white)("dd")0color(white)("dd")=0 ->(x-6)color(white)("d")=0" thus "x=+6)#

#color(blue)(x=6 and x=-2" are solutions")#