Quarter Wit, Quarter Wisdom: Inequalities with Complications - Part I

Quarter Wit, Quarter WisdomLast week we learned how to handle inequalities with many factors i.e. inequalities of the form (x – a)(x – b)(x – c)(x – d) > 0. This week, let’s see what happens in cases where the inequality is not of this form but can be manipulated and converted to this form. We will look at how to handle various complications.

Complication No. 1: (a – x)(x – b)(x – c)(x – d) > 0

We want our inequality to be of the form (x – a), not (a – x) because according to the logic we discussed last week, when x is greater than a, we want this factor to be positive. The manipulation involved is pretty simple: (a – x) = -(x – a)

So we get: – (x – a)(x – b)(x – c)(x – d) > 0

But how do we handle the negative sign in the beginning of the expression? We want the values of x for which the negative of this expression should be positive. Therefore, we basically want the value of x for which this expression itself (without the negative sign in the beginning) is negative.

We can manipulate the inequality to (x – a)(x – b)(x – c)(x – d) < 0

Or simply, multiply – (x – a)(x – b)(x – c)(x – d) > 0 by -1 on both sides. The inequality sign flips and you get (x – a)(x – b)(x – c)(x – d) < 0

e.g. Given: (4 – x)(2 – x)(-9 – x) < 0

We can re-write this as –(x – 4)(2 – x)(-9 – x) < 0

(x – 4)(x – 2)(-9 – x) < 0

-(x – 4)(x – 2)(x + 9) < 0

(x – 4)(x – 2)(x – (-9)) > 0  (multiplying both sides by -1)

Now the inequality is in the desired form.

Complication No 2: (mx – a)(x – b)(x – c)(x – d) > 0 (where m is a positive constant)

How do we bring (mx – a) to the form (x – k)? By taking m common!

(mx – a) = m(x – a/m)

The constant does not affect the sign of the expression so we don’t have to worry about it.

e.g. Given: (2x – 3)(x – 4) < 0

We can re-write this as 2(x – 3/2)(x – 4) < 0

When considering the values of x for which the expression is negative, 2 has no role to play since it is just a positive constant.

Now let’s look at a question involving both these complications.

Question 1: Find the range of x for which the given inequality holds.

-2x^3 + 17x^2 – 30x > 0

Solution:

Given: -2x^3 + 17x^2 – 30x > 0

x(-2x^2 + 17x – 30) > 0  (taking x common)

x(2x – 5)(6 – x) > 0 (factoring the quadratic)

2x(x – 5/2)(-1)(x – 6) > 0 (take 2 common)

2(x – 0)(x – 5/2)(x – 6) < 0 (multiply both sides by -1)

This inequality is in the required form. Let’s draw it on the number line.

We are looking for negative value of the expression. Look at the ranges where we have the negative sign.

The ranges where the expression gives us negative values are 5/2 < x < 6 and x < 0.

Hence, the inequality is satisfied if x lies in the range 5/2 < x < 6 or in the range x < 0.

Plug in some values lying in these ranges to confirm.

Next week, we will look at some more variations which can be brought into this form.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

9 Responses

  1. Moni says:

    Nice article, as always !

    I wanted to understand, if we have a negative constant -2 in your example of complication#2, then how will the inequality graph be plotted and what will be the range of x for the Inequality to be negative?

    e.g. In -2(x – 3/2)(x – 4) < 0
    our range for x without plotting -2 will be 3/2<x<4
    But for values of x in this range, (-2)* (x-3/2)* (x-4) will be positive right..?

    Please explain.

  2. Karishma says:

    -2(x – 3/2)(x – 4) 0

    This expression will be positive in the range x > 4 or x < 3/2. So x can take these values for -2(x – 3/2)(x – 4) to be negative.

  3. Moni says:

    Thanks!

  4. Irda says:

    Hi Karishma,

    Thanks for the explanation. I wanted to know the reason behind us looking for the negative value, as such

    When considering the values of x for which the expression is negative, 2 has no role to play since it is just a positive constant.

    Why are we solving for the negative value: is it because the expressions is less than 0.

    Thanks if you can please suggest.

    • Karishma says:

      Yes, the inequality has the ‘< 0' sign which means we need the value of x for which the inequality is negative. The constant 2 does not change sign with changing values of x so it does not affect the sign of the inequality.

  5. irda says:

    Thanks Karishma,

    Using your approach how can I solve this question:

    If x is positive, is x > 4?

    (1) (4 – x)^2 < 1

    (2) (x – 4)^2 < 1

    I couldn't come up with an approach.

    Thanks for you help

    • Karishma says:

      (1)
      (4-x)^2 < 1
      x^2 + 16 – 8x -1 < 0
      x^2 – 8x + 15 < 0
      (x – 3)(x – 5) < 0
      To be negative, x must lie between 3 and 5. Is it greater than 4? We don't know. Since x can lie anywhere between 3 and 5, it may or may not be greater than 4. Not sufficient.

      (2)
      (x – 4)^2 < 1
      This is exactly same as statement 1. Note that (4 – x)^2 = (x – 4)^2
      So this will again not be sufficient.

      Since both statements are the same and one of them is not sufficient alone, both together will also not be sufficient.
      Answer (E)

      • Catie says:

        Hi Karishma, Big fan of all your blogs! :-) Don’t you think that in the case of -2x^3 + 17x^2 – 30x > 0 3 and 20 would also work because 20-3 gives 17?

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