How do you solve #6x - 5 < 6/x#?

1 Answer
Aug 31, 2015

Answer:

#x in (-2/3, 0) uu (0, 3/2)#

Explanation:

Right from the start, you know that #x# cannot be equal to zero, since it is the denominator of a fraction. So you need to have #x!=0#.

With that in mind, multiply the left-hand side of the inequality by #1= x/x# to get rid of the denominator

#6x * x/x - 5 * x/x < 6x#

#6x^2 - 5x < 6#

Next, add #-6# to both sides of the inequality

#6x^2 - 5x - 6 < color(red)(cancel(color(black)(6))) - color(red)(cancel(color(black)(6)))#

#6x^2 - 5x - 6 < 0#

To help you determine the intervals on which this quadratic function is smaller than zero, you need to first determine its root by using the quadratic formula

#6x^2 - 5x - 6 = 0#

#x_(1,2) = (-(-5) +- sqrt((-5)^2 - 4 * 6 * (-6)))/(2 * 6)#

#x_(1,2) = (5 +- sqrt(169))/12#

#x_(1,2) = (5 +- 13)/12 = {(x_1 = (5 + 13)/12 = 3/2), (x_2 = (5 - 13)/12 = -2/3) :}#

You can thus rewrite the quadratic as

#6(x-3/2)(x+2/3) = 0#

So, you need this expression to be negative, which implies that #(x-3/2)# must be positive whenever #(x+2/3)# is negative, and vice versa.

For #x> -2/3# and #x<3/2# you get

#{(x-3/2 < 0), (x + 2/3 > 0) :} implies (x-3/2)(x+2/3) < 0#

Any value of #x>3/2# will make both terms positive, and any value of #x<-2/3# will make both terms negative. Keeping in mind that you also need #x!=0#, the solution set for this inequality will be #x in (-2/3, 0) uu (0, 3/2)#.