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 $(2,10)$

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