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Marina
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What is the value of x^3 - 2x^2 + 7? (1) 3x^3- x = 2 (2) 18 Mar 2005, 18:33
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What is the value of x^3 - 2x^2 + 7?
(1) 3x^3- x = 2
(2) x^5= 1



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What is the value of x^3 - 2x^2 + 7? (1) 3x^3- x = 2 (2) 18 Mar 2005, 22:47
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i also gueess D.
statement i, only 1 satisfies this one.
statement ii is straight.
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What is the value of x^3 - 2x^2 + 7? (1) 3x^3- x = 2 (2) 19 Mar 2005, 02:37
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banerjeea_98
I think "D", for both cases x = 1
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Banerjeea, I would appreciate if you could explain your solution.

Second is certainly a pretty straightforward case. However, in case 1, how can you get (only) x=1 from a third degree equation? I think you should have 3 roots for an equation of the third order (unless you've repeated roots, like in the second case). Because if there're other roots too, then we cannot say for sure the value of the expression, because of multiple values of x giving multiple values of the expression.

Furthermore, in arriving at x=1 as a solution, did you simply start with substitution? How did you solve the equation of the third degree?
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What is the value of x^3 - 2x^2 + 7? (1) 3x^3- x = 2 (2) 19 Mar 2005, 10:25
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kapslock
banerjeea_98
I think "D", for both cases x = 1

Banerjeea, I would appreciate if you could explain your solution.

Second is certainly a pretty straightforward case. However, in case 1, how can you get (only) x=1 from a third degree equation? I think you should have 3 roots for an equation of the third order (unless you've repeated roots, like in the second case). Because if there're other roots too, then we cannot say for sure the value of the expression, because of multiple values of x giving multiple values of the expression.

Furthermore, in arriving at x=1 as a solution, did you simply start with substitution? How did you solve the equation of the third degree?
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I used substitution for the first one, I didn't solve the third degree eqn. Wud love to know if anyone knows how to solve the 1st eqn using algebra.
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What is the value of x^3 - 2x^2 + 7? (1) 3x^3- x = 2 (2) 21 Mar 2005, 10:00
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banerjeea_98
kapslock
banerjeea_98
I think "D", for both cases x = 1

Banerjeea, I would appreciate if you could explain your solution.

Second is certainly a pretty straightforward case. However, in case 1, how can you get (only) x=1 from a third degree equation? I think you should have 3 roots for an equation of the third order (unless you've repeated roots, like in the second case). Because if there're other roots too, then we cannot say for sure the value of the expression, because of multiple values of x giving multiple values of the expression.

Furthermore, in arriving at x=1 as a solution, did you simply start with substitution? How did you solve the equation of the third degree?

I used substitution for the first one, I didn't solve the third degree eqn. Wud love to know if anyone knows how to solve the 1st eqn using algebra.
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Hmm, but isn't it a bit risky, because you don't know if there're two more roots of the equation, and since you do not know which one to use, that doesn't allow you to get to a unique solution for the expression in question?
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What is the value of x^3 - 2x^2 + 7? (1) 3x^3- x = 2 (2) 21 Mar 2005, 14:11
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D
for equation 1 i think this can work x(3x^2-1)=2
take 3x^2-1=2 gives us a possibility of only +/-1 but since x(3x^2-1)=+2
therefore ans should be 1
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What is the value of x^3 - 2x^2 + 7? (1) 3x^3- x = 2 (2) 21 Mar 2005, 16:07
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I also answered this question as D, but OA is B. I strongly believe the OA is wrong.



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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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