By the definition of the absolute value of a number, this is equivalent to stating that
x^2-9x+15 > 1x2−9x+15>1
or
x^2-9x+15 < -1x2−9x+15<−1
We will solve each inequality separately.
For the first inequality, subtracting 11 from each side, we obtain
x^2-9x+14 > 0x2−9x+14>0
=> (x-2)(x-7) > 0⇒(x−2)(x−7)>0
=> x-2 > 0⇒x−2>0 and x-7 > 0x−7>0
or
x-2 < 0x−2<0 and x-7 < 0x−7<0
=>x>7⇒x>7
or
x<2x<2
=> x in (-oo,2)uu(7,oo)⇒x∈(−∞,2)∪(7,∞)
For the second inequality, we can add 11 to both sides and use the quadratic formula to factor the left hand side.
x^2-9x+16 < 0x2−9x+16<0
=> (x-(9+sqrt(17))/2)(x-(9-sqrt(17))/2) < 0⇒(x−9+√172)(x−9−√172)<0
Because (x-(9+sqrt(17))/2) < (x-(9-sqrt(17))/2)(x−9+√172)<(x−9−√172) and the inequality holds only when one of factors is negative and the other is positive, we must have that
(x-(9+sqrt(17))/2) < 0(x−9+√172)<0 and (x-(9-sqrt(17))/2) > 0(x−9−√172)>0
=> (9-sqrt(17))/2 < x < (9+sqrt(17))/2⇒9−√172<x<9+√172
=> x in ((9-sqrt(17))/2, (9+sqrt(17))/2)⇒x∈(9−√172,9+√172)
To get our final solution set, we can take the union of both solutions we found above to obtain
x in (-oo,2) uu ((9-sqrt(17))/2, (9+sqrt(17))/2)uu(7,oo)x∈(−∞,2)∪(9−√172,9+√172)∪(7,∞)
Note that this matches what we would expect from the graph, as |x^2-9x+15|-1∣∣x2−9x+15∣∣−1 is negative on only two small intervals:
graph{|x^2-9x+15|-1 [-5.205, 14.795, -3.24, 6.76]}
