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Squares, square roots, inequalities and modulus

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eagleeye GMAT Destroyer!
28 Apr 2012
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Squares, square roots, inequalities and modulus Post Tue May 29, 2012 9:03 pm
Elapsed Time: 00:00
    It's been a month since I thought about writing the GMAT and joined this forum. This forum has been very helpful in my preparation so far, so I feel that I owe it to some of you to make your prep a little stronger.

    I see many people here struggling with concepts of square roots and modulus. So I figured I should give a little back to the community by providing some insight. I will only be talking about real numbers and up to quadratic math because that's what GMAT tests. I hope that by the time you go through this post, you will find it clearer and easier to tackle squares, moduli, and inequalities.

    Concept 1: square roots

    sqrt(real number) is never negative.

    sqrt(16) is 4. (It is never -4).
    sqrt(0) is 0.
    sqrt(9) is 3. etc.

    When solving questions, especially DS type ones, you should always know that sqrt(x) >=0.

    Concept 2: Squares

    Look at the two equations below:

    x^2 = 16
    x = sqrt(16)

    They are NOT the same equations.

    x = sqrt(16) is a linear equation that has only one answer, which is x=4.
    x^2=16 is a quadratic equation which has two roots, and hence two answers x= +4 and x=-4.

    Concept 3: Modulus

    Modulus is always non-negative. |x| >=0.

    To solve modulus questions, especially inequalities, use the following.

    if x<=0, |x| = -x (This is very important).
    if x>=0, |x| = x (This is equally important).

    Since x can be either positive or negative, if no other information is given,
    |x| = 5 has two solutions, x = +5, and x = -5.
    However x = |5| has only one solution which is x = 5.

    you may recognize that this is similar to the square root, square statements I provided earlier. In fact, we can combine the two as I have at the end of this article.

    So if x = -5; |x| = -x = -(-5) , hence |x| = 5 which is not negative.
    If x=0; |x| = 0
    If x = 3; |x| = x , hence |x| = 3, which is non-negative as well.

    Concept 4: Inequalities.

    The first thing to notice is whether it is a strict inequality or not. Noticing it can mean the difference between a right and a wrong answer.

    1). x > 0 is a strict inequality , X will always be positive. Therefore, x can't be negative. Also, x can't be 0.

    2). x>=0, can mean two things, and ONLY one of those two things can be true.
    a. Either x = 0, or
    b. x > 0.

    So when you are solving for DS type questions, always see if the says a>b or a>=b.

    3) If something is > 0, it is said to be positive.
    But if something is >=0 , it is said to be non-negative. Pay attention to the language of the question.

    4). You can ALWAYS add or subtract numbers to an inequality, without changing the sign.
    Best to illustrate with an example.
    We are given x-5 < 9; we can easily add 5 to both sides or subtract 9 from both sides, and the inequality will still be correct. so x-5<9 is same as x-5+5 < 9+5 which means x < 14.

    5) More importantly, you CANNOT just multiply or divide a variable in an inequality blindly.
    . To do so, keep the following in mind when working with fractional inequalities.

    a. If x/y > 6 and you want to bring y to the denominator, remember this.
    If y is negative, and you multiply both sides with y the sign will be reversed. (Very very important)
    So, in this case x/y < 6 will become x > 6y.

    b. The opposite is also true.
    If y is positive, and you multiply both sides with y, the sign remains the same. (Again important)

    If you are doing a DS type question, and you don't know whether the denominator is positive or negative, solve for separate cases where y>0, and y<0.

    Concept 6: Final concept.
    A frequently repeated concept that I see my fellow BTGers struggle with is modulus or squares combined with inequalities.

    Let's look at an example.

    7/|x-a| < 13. where x is not equal to a.
    Let's use the concepts above. First |x-a|>=0, Since x is not equal to a, it reduces to |x-a|>0.
    Now since we know that the denominator is positive, we can multiply both sides with |x-a| without changing the sign. Then, we get 7 < 13*|x-a| or |x-a| > 7/13.

    To solve this, remember that |x-a| is either (x-a) or -(x-a) depending on whether it is positive or negative.
    so if x-a is positive, |x-a| = x-a ; we get x-a>7/13 or x > a + 7/13
    if x-a is negative , |x-a| = -(x-a); we get -(x-a) > 7/13 which becomes x-a<-7/13 which gives us the other value of x as x < a-7/13. Hence x > a+7/13 or x < a-7/13.

    Try solving |x-a|<b to give a-b <x <a+b. Then commit it to heart.

    Another concept that can be derived from the ones at the start is that sqrt(x^2) = |x|.

    Even if all this information may appear very basic, it is very powerful, you need to have these concepts clear as glass to tackle the GMAT in the short time provided.

    Good luck with your preparations. I sincerely hope that we all do well.

    Let me know if this helps Smile

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    GMAT Kolaveri Really wants to Beat The GMAT!
    23 Mar 2012
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    Post Wed May 30, 2012 7:23 am
    1. Only linear equations in x (degree of x is 1) have unique solution for x.
    non-linear equations (x^2,x^3) have more than one solution.

    2. |x| < a implies -a < x < a

    3. Do not assume any variable to be an integer unless it is explicitly stated in the question.

    4. Use extreme values when testing the statements with numbers.
    suppose x < 2. we can test with x =1.9999
    X < 3.45, we can use X = 3.4444

    5. When X is mentioned as integer, always test with 0

    6. align inequalities having the same sign so that they can be visually compared.

    Regards and Thanks,
    Vinoth@GMAT Kolaveri

    Click the thank you button if you like my reply Smile

    Thanked by: das.ashmita

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