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x^2 + 1

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x^2 + 1 [#permalink] New post 10 Sep 2017, 07:52
Expert's post
00:00

Question Stats:

71% (01:01) correct 28% (00:29) wrong based on 28 sessions
Quantity A
Quantity B
\(x^2 + 1\)
\(2x - 1\)



A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


Practice Questions
Question: 8
Page: 115
Difficulty: medium
[Reveal] Spoiler: OA

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Re: x^2 + 1 [#permalink] New post 10 Sep 2017, 07:58
Expert's post
Explanation

Starting from quantity A we do know that any value we substitute in X will always positive due to its square. On the other hand quanity B coulb be also a negative number

Moreover, we should take in account the value could be a fraction such as \(\frac{1}{2}\) or an integer such as 5

A) \(\frac{1}{2^2}\) + 1 = 1.25

Or \(5^2\) + 1 = 26

B) 2* \(\frac{1}{2}\) - 1 = 0

Or using 5 we do have 10 - 1 = 9

Or -5 we do have -10 - 1 =-11

Quantity A is always greater.

As a side note, the OE is too cumbersome

an excerpt

Quote:
The left-hand side of the comparison is the square of a number. Since the square of a number is always greater than or equal to 0, and 0 is greater than the −1, the simplified comparison is the inequality and the resulting 2 (x − 1) > −1 the relationship is greater than (>). In reverse order, each simplification step implies the inequality greater than (>) in the preceding comparison. Therefore, Quantity A is greater than Quantity B.


As you can see is difficult. Instead, stick with the basic. In this case, picking numbers and find the fastest solution within 30 seconds and move on.
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Re: x^2 + 1 [#permalink] New post 12 Sep 2017, 07:32
Again let A = B

x^2 + 1 = 2x - 1

x^2 - 2x = -2

The RHS is a parabola whose vertex has a minimum. You can take the derivative of it (2x - 2) to get the lowest x-coordinate (-1) and then plug that into the actual equation to get the minimum y value, -1. Since -2 is always smaller the answer is A. Taking the derivative might be out of the scope of the GRE but honestly it's such a useful thing to know for these sort of questions. I don't particularly like the plug values method because you might miss some numbers (people tend to only use integers even if that's not mentioned in the Q) or make some silly mistake. If you do end up using number it's best to make a table and carefully solve questions so as to not make silly mistakes.
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Re: x^2 + 1 [#permalink] New post 12 Sep 2017, 08:36
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Expert's post
Carcass wrote:
Quantity A
Quantity B
\(x^2\) + 1
2x - 1



Another approach...

ASIDE: I know that x² - 2x + 1 is a perfect square that can be factored.
So, let's create that situation by performing matching operations

Given:
Quantity A: x² + 1
Quantity B: 2x - 1

Subtract 2x from both quantities:
Quantity A: x² - 2x + 1
Quantity B: -1

Factor Quantity A:
Quantity A: (x - 1)²
Quantity B: -1

We know that (something)² will ALWAYS evaluate to be greater than or equal to zero
As such, Quantity A will ALWAYS be greater than -1 (Quantity B)

Answer:
[Reveal] Spoiler:
A


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Re: x^2 + 1   [#permalink] 12 Sep 2017, 08:36
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