Absolute value concept

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Absolute value concept

by abhirup1711 » Fri Apr 19, 2013 2:10 am
Is root x a prime number?

(1)|3x-7| = 2x+2

(2) x^2 = 9x

The answer is (c) and the solution comes out by considering x>= 7/3 and x <=7/3

On the other hand in another sum:

Is x > 0?

(1) |x + 3| = 4x - 3

(2) |x + 1| = 2x - 1

Where, the answer is (d) and the solution comes out by just considering RHS has to be +ve because LHS is absolute value.

May I ask an expert to please tell me when do I consider which of the two cases and solve a sum?

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by Anju@Gurome » Fri Apr 19, 2013 2:40 am
abhirup1711 wrote:May I ask an expert to please tell me when do I consider which of the two cases and solve a sum?
In the first case we need to know the exact value of x in order to determine whether its square root is prime or not. So, we need to solve the equation to pinpoint the possible values of x.

Whereas in the second case, we only need to determine whether x is positive or negative. So, we may not need to solve the equation.

But when to solve absolute value equation/inequalities and when not - ultimately it depends upon your practice and experience.
For example, in the second case, an experienced student will be able to identify that we actually do not need to solve the equation as
  • In statement 1, (4x - 3) ≥ 0 as it is equal to an absolute value.
    So, x ≥ 3/4 > 0 ---> x > 0

    In statement 2, (2x - 1) ≥ 0 as it is equal to an absolute value.
    So, x ≥ 1/2 > 0 ---> x > 0
But do not take this as the ultimate trick.
If the problem was as follows,
Is x > 0?

(1) |x + 3| = 4x - 3
(2) |x + 2| = 2x + 1
Then we can easily say statement 1 is sufficient as per our earlier analysis.

For statement 2, using the same trick, (2x + 1) ≥ 0
So, x ≥ -1/2

Now, one may conclude that as x is greater than or equal to -1/2, x can be either positive or negative and statement 2 is insufficient.
But that will be a wrong conclusion.

Here is why...
If we solve the inequality in this case as follows,
  • # If x ≥ -2 ---> |x + 2| = (2x + 1)
    So, (x + 2) = (2x + 1)
    --> x = 1

    # If x < -2 ---> |x + 2| = -(x + 2)
    So, -(x + 2) = (2x + 1)
    --> x = -1
    Now, -1 is not less than -2 as we initially assumed.

    Hence, our only possible solution is x = 1, i.e. x > 0
    So, statement 2 is also sufficient.
To sum it up, tricky methods are less time consuming and should be tried first, but there is no guarantee that a tricky method will always lead you to the correct answer. This is because tricky methods are often problem-specific, i.e. a tricky method which is applicable to problem A may not be applicable to problem B even if both A and B are on the same concept.

Solve more and more problems with different approaches and you will automatically learn when to use which method.

My general tip for solving absolute value problems will be : Before jumping in to solve the equation/inequality, think in terms of distance on the number line...
  • 1. Does it solve the problem? If yes, we are done.
    2. If no, solve the equation/inequality.
Hope that helps.
Anju Agarwal
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by abhirup1711 » Fri Apr 19, 2013 4:06 am
Thanks a lot Anju. Much appreciated.