Question #f4b24

1 Answer
Feb 13, 2017

Answer:

#3 < x < 6# or #-2 < x < -1#.

Explanation:

First, #x != 0# since we need the denominator to be non-zero. Assuming that #x > 0#, we can multiply by #x# without switching the inequality:

#x < x^2 - 6 < 5x => x < x^2 - 6# and #x^2 - 6 < 5x#.

The first one can be rewritten as so, by subtracting #x#:

#0 < x^2 - x - 6 => 0 < (x-3)(x+2) => x > 3# or #x<-2#.

However, we took #x > 0#, so only #x > 3# is accepted.

The second inequality, is written as:

#x^2 - 5x - 6 < 0 => (x-6)(x+1) < 0 => -1 < x < 6#

Once again, since #x >0#, we accept #0< x < 6#.

If #x > 3# and #0 < x < 6#, then #3 < x < 6#.

Now, let #x < 0#. Then #x > x^2 - 6 > 5x#, since we multiplied by a negative number (#x#), so the inequality is now the other way around. We can solve them the same way, remembering now that the #># changes into #<# and the #<# into a #>#. We arrive at the results:

#0 > x^2 - x - 6 => 0 > (x-3)(x+2) => -2 < x < 3#

This time, #x < 0# so we accept #-2 < x < 0#.

The second inequality is:

#x^2 - 5x - 6 > 0 => (x-6)(x+1) > 0 => x > 6# or #x < -1#

Since #x < 0#, we accept #x < -1#.

Now, we can combine the inequalities:

#-2 < x < 3# and #x < -1# means #-2 < x < -1#

So in the end:

#3 < x < 6# or #-2 < x < -1#.

Or in interval notation:

#x in (-2, -1)uu(3,6)#


Note that you don't have to factor the quadratic polynomials there to solve the quadratic inequality. Simply use the quadratic formula to find the roots, and then if #r_1,r_2# are these roots, (#r_1 < r_2#) then:

The polynomial is positive, if #x > r_2# or #x < r_1#
The polynomial is negative, if #r_1 < x < r_2#

If the coefficient of #x^2# is negative, the above are reversed

And finally, the quadratic formula:

#r_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a)#

where #a,b,c# the coefficients of the polynomial from highest to lowest degree.