MGMAT: Inequalities & Absolute Value (why did A.V. pop u

[ostrowskiamy]

Hi folks! I ran into another tough spot during a seemingly-simple inequality question:

(x+1)(^2) < 36
sq.root of this is < 36
I x +1 I < 6
-7 < x < 5

I mostly understand why I need to put the absolute value brackets in, during the third step...but I think that's mostly because I know I'm working with absolute values & just studied that section of the book! How can I remember to do that on test day? Should I ALWAYS do that when working with inequalities, where part of the solution (here, the variable) is in parentheses?

Thanks again!

Best,
Amy

[mkdureja]

Reason why you need to use absolute values in this case is:

For any negative number, its square is +ve, eg. 5² = -5² = 25
In solving any problem, remember if x² = 25 , then x can be either +5 or -5.
Of course, in some problems, like if problem asks you to say, number of people, which can't be a -ve value, you can ignore the negative value in that case, but in other cases, be it inequality problem, or any other problem, you have to consider the negative value.

Take the case of cubes, cube of a -ve number is -ve., ie. 5³ = 125. and -5³ = -125.
So, if the above problem would've been (x+1)³<125, you dont need to use absolute value in this case.,
answer of this would be x+1<5, or x<4.

When dealing with even exponents, you have to use absolute value brackets.

Hope its clear to you.

[Anurag@Gurome]

I mostly understand why I need to put the absolute value brackets in, during the third step...but I think that's mostly because I know I'm working with absolute values & just studied that section of the book! How can I remember to do that on test day?

Just remember that by definition square root of any algebraic quantity x² is |x|
Hence, square root of (x + 1)² is |x + 1|, square root of (x + y + z + 2xy + abc)² i |x + y + z + 2xy + abc| etc.

For a more detailed explanation, refer to the post here >> https://www.beatthegmat.com/contradictor ... tml#476587

[Anurag@Gurome]

Hi folks! I ran into another tough spot during a seemingly-simple inequality question:

(x+1)(^2) < 36

There is another way to solve this problem without bringing absolute values if you're not confident with them.

--> (x + 1)² < 36
--> (x + 1)² - 6² < 0
--> (x + 1 - 6)(x + 1 + 6) < 0
--> (x - 5)(x + 7) < 0

Hence, (x - 5) and (x + 7) are of opposite signs.
Hence, the following two cases are possible...

• 1. (x - 5) < 0 and (x + 7) > 0 ---> x < 5 and x > -7 ---> -7 < x < 5
  2. (x - 5) > 0 and (x + 7) < 0 ---> x > 5 and x < -7 --> Not possible

Hence, -7 < x < 5