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.

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

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

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

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