How do you factor and solve #x^2+x-6=0#?

1 Answer
Aug 1, 2016

Answer:

#x=2# and #x=-3#

Explanation:

In a quadratic equation #ax^2+bx+c=0#, to solve one splits the middle term in two parts so that their sum is #b# and product is #ac#.

Hence, in #x^2+x-6=0# one needs to split #1xx(-6)=-6# in two parts whose sum is #1#. It is apparent that these are #3# and #-2#, hence #x^2+x-6=0# can be written as

#x^2+3x-2x-6=0# or

#x(x+3)-2(x+3)=0# or

#(x-2)(x+3)=0#

i.e. either #x-2=0# i.e. #x=2#

or #x+3=0# i.e. #x=-3#