How do you solve #p^{2} - 4p = 12#?

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Answer 1 · 2016-09-24T11:03:02

#x=4 or x = -2#

It is a quadratic - make it equal to 0.

#p^2 -4p -12 = 0" "larr# find factors

Find factors of 12 which DIFFER (because of MINUS 12) by 4.

The factors are 6 and 2. [ 6x2=12 and 6-2 =4]

#(x-4)(x+2) =0" "larr# there must be more negatives. (-4)

One of the factors must be 0

If #x-4 = 0, rarr x = 4#
If #x+2 =0, rarr x = -2#