How do you solve the inequality: #|x^2-9x+15|>1 #?

1 Answer
Apr 19, 2016

Answer:

#x in (-oo,2) uu ((9-sqrt(17))/2, (9+sqrt(17))/2)uu(7,oo)#

Explanation:

By the definition of the absolute value of a number, this is equivalent to stating that

#x^2-9x+15 > 1#
or
#x^2-9x+15 < -1#

We will solve each inequality separately.


For the first inequality, subtracting #1# from each side, we obtain

#x^2-9x+14 > 0#

#=> (x-2)(x-7) > 0#

#=> x-2 > 0# and #x-7 > 0#
or
#x-2 < 0# and #x-7 < 0#

#=>x>7#
or
#x<2#

#=> x in (-oo,2)uu(7,oo)#


For the second inequality, we can add #1# to both sides and use the quadratic formula to factor the left hand side.

#x^2-9x+16 < 0#

#=> (x-(9+sqrt(17))/2)(x-(9-sqrt(17))/2) < 0#

Because #(x-(9+sqrt(17))/2) < (x-(9-sqrt(17))/2)# 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# and #(x-(9-sqrt(17))/2) > 0#

#=> (9-sqrt(17))/2 < x < (9+sqrt(17))/2#

#=> x in ((9-sqrt(17))/2, (9+sqrt(17))/2)#


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)#

Note that this matches what we would expect from the graph, as #|x^2-9x+15|-1# is negative on only two small intervals:

graph{|x^2-9x+15|-1 [-5.205, 14.795, -3.24, 6.76]}