How do you solve x^2-2x-12=0 algebraically?

1 Answer
Apr 5, 2016

x~~4.606" " -2.606 to 3 decimal places

color(green)(x=1+-sqrt(13))

Explanation:

The whole number factors of 12 are {6,2}, {3,4} and none of these have a difference of 2 ( from 2x). So it must have none integer solutions for x. Thus we need to use the formulas. These formula are something worth remembering.

It takes quite a bit of work to make them stick!

Standard equation form:" "y=ax^2 +bx+c

where color(magenta)(x=(-b+-sqrt(b^2-4ac))/(2a)) ............................(1)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Another one worth remembering is the completed square format.
Otherwise known as Vertex Form.

" "color(magenta)(y=a(x+b/(2a))^2+c -[(b/2)^2]) .............................(2)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Using equation (1)")

=>x" "=" "(-(-2)+-sqrt( (-2)^2-4(1)(-12)))/(2(1))

=>x" "=" "(2+-sqrt( 52))/2

=>x" "=" "(2+-sqrt(2^2xx13))/2

color(green)(=>x= 1+-sqrt(13))
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Using equation (2)")

y=1(x-1)^2-12-1

y=(x-1)^2-13

Set y=0

0=(x-1)^2-13

Add 13 to both sides

13=(x-1)^2

Square root both sides

+-sqrt(13)=x-1

Add 1 to both sides

color(green)(x=1+-sqrt(13))
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
x~~4.606" " -2.606 to 3 decimal places