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

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Answer 1 · 2015-05-19T07:58:45

We can Split the Middle Term of this equation and factorise it to find the solution.

In this technique, if we have to factorise an expression like #ax^2 +bx + c#, we need to think of two numbers such that:

#N_1 xx N_2 = a xx c = 1 xx (-12) = -12#

and

#N_1 +N_2 = b = -1#

After trying out a few numbers we get #N_1 = -4# and #N_2 =3#
#(-4) xx 3 = -12#, and #(-4) + 3= -1#

# x^2 -x -12 = x^2 -4x +3x -12 #
#= x(x-4)+3 (x-4)#
#= (x+3)(x-4)#

the solution is #color(blue)x=-3 , color(blue)x=4#