Question

Find the complete range of values for x that satisfy |x + 3| > x^2-11x+30

Is is (- infinity,3) U (9,infinity)

Answer 1

two cases
Case 1:
x+3>x^2-11x+30
0>x^2-12x+27
0>(x-3)(x-9)
this is only true when one of the two things you're multiplying is positive and one is negative. This happens between 3 and 9 (but not including these two numbers), so
D:(3,9)
Case 2:
-(x+3)>x^2-11x+30
x+3<-x^2+11x-30
x^2-10x+33<0
quadratic here
10+/-sqrt(100-4(33)) all over 2
10+/-sqrt of a negative. We can stop considering this one, unless you want to involve imaginary roots.
This means it is only true for the answers from case 1: (3,9)

Answer 2

|x + 3| > (x - 5)(x-6)

Now

x + 3 > (x-5)(x-6) or x + 3 < -(x - 5)(x - 6)

x + 3 - x^2 +11x -30 >0 or x + 3 +x^2 -11x + 30 <-

-x^2 + 12x -27 >0 or x^2 - 10x +33<0

(x -9)(x -3) < 0 or Now b^2 - 4ac < 0 so roots don't exist (real)




3 9
---------------|--------------------------|-------
+++ ---- +++

So Hence Solution is From (3,9)

Answer 3

two good answers! mine was 1 minute faster!!