Hello Pros,
My questions is regarding statement 1
What is the value of 5x^2 +1?
1) x^3 - x^2 - 9x + 9 = 0
my approach yielded that it is sufficient; however the OA says that it is not. Here is what I did
x^2(x-1) -9(x-1) = 0
x^2(x-1) = 9(x-1)
divide both sides by x-1
x^2 = 9 hence 5x^2 +1 = 46
the OA explanation was that it did not divide both sides by x-1 as i did
however, it factored further the expression to get (x^2 - 9)(x-1)(x-1) = 0
hence x can be 3, -3 or 1
my question is what did i do wrong to eliminate that x could be 1
My factorization looked fine to me until i saw this explanation!
pls advise.
factorization
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When you divide by an algebraic expression (like x-1), there's a chance that you are dividing by zero.
If you divide by zero, the resulting equation may be meaningless.
For example, we know that (2)(0) = (1)(0)
If we divide both sides by zero, we can't conclude that 2 = 1
The same applies to x^2(x-1) = 9(x-1)
If we divide both sides by (x-1), we can't conclude that x^2 = 9
Cheers,
Brent
If you divide by zero, the resulting equation may be meaningless.
For example, we know that (2)(0) = (1)(0)
If we divide both sides by zero, we can't conclude that 2 = 1
The same applies to x^2(x-1) = 9(x-1)
If we divide both sides by (x-1), we can't conclude that x^2 = 9
Cheers,
Brent
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Hi Amrabdelnaby,
You have to be VERY careful when you divide by a variable. By doing so, you might end up 'losing' one of the solutions to the original equation/question.
Here's a simple example:
X^2 = X
Mathematically, this equation has 2 solutions: 0 and 1. You can prove that by doing the following....
X^2 = X
X^2 - X = 0
X(X-1) = 0
X = 0, 1
However, if you chose to divide both sides by X to start, you'd have.....
X^2 = X
X = 1
By dividing, you "lost" the solution X=0.
These types of 'tests' often show up in DS questions (since DS questions test the 'thoroughness' of your thinking, among other things). Thus, it's important to be careful about how you choose to do your math - the scope of the correct answer depends on it.
GMAT assassins aren't born, they're made,
Rich
You have to be VERY careful when you divide by a variable. By doing so, you might end up 'losing' one of the solutions to the original equation/question.
Here's a simple example:
X^2 = X
Mathematically, this equation has 2 solutions: 0 and 1. You can prove that by doing the following....
X^2 = X
X^2 - X = 0
X(X-1) = 0
X = 0, 1
However, if you chose to divide both sides by X to start, you'd have.....
X^2 = X
X = 1
By dividing, you "lost" the solution X=0.
These types of 'tests' often show up in DS questions (since DS questions test the 'thoroughness' of your thinking, among other things). Thus, it's important to be careful about how you choose to do your math - the scope of the correct answer depends on it.
GMAT assassins aren't born, they're made,
Rich
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hiAmrabdelnaby wrote:Hello Pros,
My questions is regarding statement 1
What is the value of 5x^2 +1?
1) x^3 - x^2 - 9x + 9 = 0
my approach yielded that it is sufficient; however the OA says that it is not. Here is what I did
x^2(x-1) -9(x-1) = 0
x^2(x-1) = 9(x-1)
divide both sides by x-1
x^2 = 9 hence 5x^2 +1 = 46
the OA explanation was that it did not divide both sides by x-1 as i did
however, it factored further the expression to get (x^2 - 9)(x-1)(x-1) = 0
hence x can be 3, -3 or 1
my question is what did i do wrong to eliminate that x could be 1
My factorization looked fine to me until i saw this explanation!
pls advise.
there are already explanations given that we lose out on the value of x as 0..
where did you go wrong in factorization was from the step...
x^2(x-1) -9(x-1) = 0..
this is equal to
(x^2-9)(x-1) = 0...
so we get the values 3,1 and -3