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kaplan ds test 1 Q20. value of x(square) - 1. OA wrong ?

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kaplan ds test 1 Q20. value of x(square) - 1. OA wrong ?

by raghavsarathy » Thu Jun 25, 2009 6:01 pm
Question - What is the value of x(square) - 1

1. Value of x(square) - 5x= -6
2. (x(square) - 1 )/(x-1) = 1

[spoiler]OA = B[/spoiler]

[spoiler]IMO = C[/spoiler]

Each of the 2 quadratic eqns would give us 2 values of x and hence we will find the comon value of x using both eqns and hence my ans was C

The problem with the Kaplan ans is that for eqn 2 , they cancel out (x-1) after expanding the numerator. Isn't this wrong ? because we are in essence losing out a value of x. [spoiler][/spoiler]
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Re: kaplan ds test 1 Q20. value of x(square) - 1. OA wrong ?

by ssmiles08 » Thu Jun 25, 2009 6:08 pm
raghavsarathy wrote:Question - What is the value of x(square) - 1

1. Value of x(square) - 5x= -6
2. (x(square) - 1 )/(x-1) = 1

[spoiler]OA = B[/spoiler]

[spoiler]IMO = C[/spoiler]

Each of the 2 quadratic eqns would give us 2 values of x and hence we will find the comon value of x using both eqns and hence my ans was C

The problem with the Kaplan ans is that for eqn 2 , they cancel out (x-1) after expanding the numerator. Isn't this wrong ? because we are in essence losing out a value of x. [spoiler][/spoiler]
No I dont think Kaplan is wrong in canceling out the numerator, b/c if you plug the value of x after cancelling out the numerator, it still gives you a value of 1.

x^2 - 1 is essentially the same thing as (x^2 - 1^2)/(x-1) = 1

[(x+1)(x-1)]/[x-1] = 1

x+1 = 1

x = 0.

you can plug back 0 in the original case 2 and see that, it is the only value of x that works out to be 1.

so (B)
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by Domnu » Thu Jun 25, 2009 6:11 pm
In part 2, canceling this out is okay, because the rule is: you can't ever divide or multiply by 0. Cancelling out the x - 1 is okay since we know that x can't ever be 1 in the second equation (see what happens if you try substitution). If we lose a value in the second equation, there's only one value left to try, so we are done. The answer is thus B.
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by raghavsarathy » Fri Jun 26, 2009 10:25 am
Thanks guys. i missed the point that for the second Eqn x=1 would result in an invalid eqn with denomnator 0.
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by rah_pandey » Fri Jun 26, 2009 10:26 pm
As far as I remember from high school maths x<>1 should be specifically mentioned in order for cancellation to take place. we cannot assume that x<>1 because it would make the equation invalid. I think Stuart can explain more on this.
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