Question

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

Answer 1

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

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

Explanation:

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")`