How do you solve #x^2 – 12x + 32 <12#?

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Answer 1 · 2017-12-01T01:25:49

Start with equality, find the region and check where it is valid

We have the inequality as
#x^2-12x+32 < 12#
#x^2-12x+20 < 0#

Start with equality

#x^2-12x+20 = 0#
#(x-10)(x-2) = 0#
#x = 2,10#

So we have 3 ranges.
#(-\infty,2), (2,10), (10,\infty)#

Let us take some number from each region and check where it is less than 0.

#x=0# then we have #0^2-12\times 0+20 = 20 > 0#
#x = 11# then we have #11^2-12\times 11+20 = 9 >0#
#x = 3# then we have #3^2-12\times 3+20 = -7<0#

Hence the region where the inequality holds is #(2,10)#