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

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Answer 1 · 2015-02-03T13:58:15

We can use the Sum-Product method.

Bring everything to one side:
#x^2-x=12->x^2-x-12=0#

We now have a quadratic equation of the form
#ax^2+bx+c=0# where #a=1, b=-1 and c=-12#

We find two numbers that will give #c=-12# as a product and #b=-1# as a sum (or difference).
We can try #1*12, 2*6,3*4#

#3and4# will fit, with a #-#sign to the #4#, as that would make the sum #3-4=-1# and the product #3*-4=-12#

Now we can rewrite the equation:
#(x+3)(x-4)=0#

Which leaves us with two possibilities:
#(x+3)=0->x=-3# OR #(x-4)=0->x=4#

Answer:
#x=-3 or x=4#