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Need expert help - 2 different answer with 2 diff methods

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Need expert help - 2 different answer with 2 diff methods

by voodoo_child » Wed Sep 21, 2011 8:14 am
If x is a positive integer, is sqrt(x) < 2.5x - 5 ?

1. x < 3
2. x is a prime number

There are two ways to approach this problem - substitute the value of x or solve the equation first.

The first method works fine.

However, if I square the equation,

x < 6.25 (x^2 - 4x + 4)

Hence, 6.25x^2 - 26x + 25 > 0 must be satisfied.

However, if you substitute x = 2 => LHS > 0 Satisfied.
If you substitute x =1 => LHS < 0 not satisfied. Can anyone tell me what's wrong with the second method?

Thanks

Voodoo Child
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by shankar.ashwin » Wed Sep 21, 2011 9:15 am
IMO A

You have calculated it wrong I guess;

If x=2; you get
6.25(4) - 26(2) + 25 = -2 which is still lesser than 0.
voodoo_child wrote:If x is a positive integer, is sqrt(x) < 2.5x - 5 ?

1. x < 3
2. x is a prime number

There are two ways to approach this problem - substitute the value of x or solve the equation first.

The first method works fine.

However, if I square the equation,

x < 6.25 (x^2 - 4x + 4)

Hence, 6.25x^2 - 26x + 25 > 0 must be satisfied.

However, if you substitute x = 2 => LHS > 0 Satisfied.
If you substitute x =1 => LHS < 0 not satisfied. Can anyone tell me what's wrong with the second method?

Thanks

Voodoo Child
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by cans » Wed Sep 21, 2011 9:16 am
voodoo_child wrote:If x is a positive integer, is sqrt(x) < 2.5x - 5 ?

1. x < 3
2. x is a prime number

There are two ways to approach this problem - substitute the value of x or solve the equation first.

The first method works fine.

However, if I square the equation,

x < 6.25 (x^2 - 4x + 4)

Hence, 6.25x^2 - 26x + 25 > 0 must be satisfied.

However, if you substitute x = 2 => LHS > 0 Satisfied.
If you substitute x =1 => LHS < 0 not satisfied. Can anyone tell me what's wrong with the second method?

Thanks

Voodoo Child
Squaring with equal to sign is simple. lets say a=b then a^2=b^2
But with <, > sign its not that simple...
-2>-4 but when you square, 4>16 is not true...
Thus when you square sqrt(x) < 2.5x-5, you must consider 2.5x-5>=0 or x>=2.
That's why when you chose x=2, it worked and when x=1, it didn't..
I hope its clear....
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Cans!!
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