Inequalities: 0 <= 8-1/5x < 5 and 2x(9x+8) < (3x-1)(6x+1)


New member
Jan 11, 2008
Our teacher gave us some practice problems for our midterm and I can't figure out two of them. I don't understand how to get the right answer. I understand parts of it, which I will show.

The first problem is:


here is how I solved it

answer (40,15]

I know my notations are right, but I can't figure out the numbers

The second problem is


Here is how I solved it

answer (-1/21, Infinity)

once again i know my notations are right but I can't get the numbers.


Super Moderator
Staff member
Feb 4, 2004
gierhame said:

answer (40,15]
Um... Are you sure that forty is less than fifteen...? :oops:

Your formatting is ambiguous; I will guess that you mean the following:

. . . . .\(\displaystyle 0\, \leq\, 8\, -\, \frac{1}{5}x\, <\, 5\)

You then multiplied through by 5 to get:

. . . . .\(\displaystyle 0\, \leq\, 40\, -\, x\, <\, 25\)

I'm not sure what happened after that...? But a good next step would be to subtract forty from all three "sides":

. . . . .\(\displaystyle -40\, \leq\, -x\, <\, -15\)

Since -1 is negative, you will of course need to reverse the inequalities when you divide through to complete the exercise.

gierhame said:

answer (-1/21, Infinity)
You're close, but you made an arithmetic and a direction error.

In this specific instance, the squared variable terms cancel out, and you end up with a linear inequality, which makes things fairly simple. Usually, however, you'll have to be more careful. And since you don't know the sign on x, you cannot multiply through by any term containing x (since you don't know if you need to flip the inequality sign, should that factor turn out to be negative).

I'm not sure what you did...? One way to solve quadratic inequalities is to get everything on one side of the inequality symbol, and solve the related quadratic equation for the zeroes (that is, for where the quadratic crosses the x-axis). These zeroes divide the number line (the x-axis) into intervals. The associated parabola is either above or below the x-axis on each interval. If you need the quadratic to be positive, then you take the interval(s) where the parabola is above the axis; otherwise, you take the interval(s) below.

For instance, given:

. . . . .\(\displaystyle x^2\, -\, 4x\, -\, 12\, >\, 0\) would solve the associated equality:

. . . . .\(\displaystyle x^2\, -\, 4x\, -\, 12\, =\, 0\)

. . . . .\(\displaystyle (x\, -\, 6)(x\, +\, 2)\, =\, 0\)

. . . . .\(\displaystyle x\, =\, 6\, \mbox{ or }\, x\, =\, -2\)

Then your intervals would be \(\displaystyle (-\infty,\, -2)\), \(\displaystyle (-2,\, 6)\), and \(\displaystyle (6, \infty)\). Since the quadratic is positive, then the parabola is upward-opening. Thus, it will be above the x-axis on the ends, and below it in the middle. The solution is then:

. . . . .\(\displaystyle x\, \in\, (-\infty,\, -2)\, \cup\, (6, \infty)\)

Had the inequality be an "or equal to" one, the endpoints (in this case, -2 and 6) would also have been included in the solution.

Have fun! :D