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Re: I got this wrong on the PowerPrep! help? [#permalink]
03 Jun 2016, 04:01

1

This post received KUDOS

Expert's post

rbtta23 wrote:

If x is an integer, which of the following must be an even integer? A: x^2-x-1 B: x^2-4x+6 C: x^2-5x+5 D: x^2+3x+8 E: x^2+2x+10

We first must recognize that when squaring an odd integer, the result will still be odd and when squaring an even integer, the result will still be even. That fact will help when analyzing the answer choices.

It will also be helpful to use our addition, subtraction, and multiplication rules, for odd and even numbers, when analyzing the answer choices.

Addition Rules

E + E = Even O + O = Even E + O = Odd

*the same rules apply for subtraction

Multiplication Rules

E x E = Even E x O = Even O x O = Odd

Now let's analyze each answer choice for when x is odd and when x is even.

A) x^2-x-1

When x is even: E - E - 1 = odd

Since we have found an odd result, answer A is not correct.

B) x^2 - 4x + 6

When x is even: E - E + 6 = even

When x is odd: O - E + 6 = odd

Since we have found an odd result, answer B is not correct.

C) x^2 - 5x + 5

When x is even: E - E + 5 = odd

Since we have found an odd result, answer C is not correct.

D) x^2 + 3x + 8

When x is even: E - E + 8 = even

When x is odd: O + O + 8 = even

Since we only get an even result, answer D is correct.

E) x^2 + 2x + 10

When x is even: E + E + 10 = even

When x is odd: O + E + 10 = odd

Since we have found an odd result, answer E is not correct.

Re: I got this wrong on the PowerPrep! help? [#permalink]
09 Jun 2016, 09:52

2

This post received KUDOS

Expert's post

rbtta23 wrote:

If x is an integer, which of the following must be an even integer? A: x² - x - 1 B: x² - 4x + 6 C: x² - 5x + 5 D: x² + 3x + 8 E: x² + 2x + 10

You can also try plugging in an odd integer and an even integer to see what happens.

Start by plugging in an EVEN integer. The easiest even value to plug in is x = 0 We get... A: 0² - (0) - 1 = -1 = ODD (ELIMINATE) B: 0² - 4(0) + 6 = 6 = EVEN (KEEP) C: 0² - 5(0) + 5 = 5 = ODD (ELIMINATE) D: 0² + 3(0) + 8 = 8 = EVEN (KEEP) E: 0² + 2(0) + 10 = 10 = EVEN (KEEP)

We're down to B, D and E Now plug in an ODD integer. The easiest odd value to plug in is x = 1 B: 1² - 4(1) + 6 = 3 = ODD (ELIMINATE) D: 1² + 3(1) + 8 = -1 = EVEN (KEEP) E: 1² + 2(1) + 10 = 13 = ODD (ELIMINATE)

Brent Hanneson – Creator of greenlighttestprep.com If you enjoy my solutions, you'll like my GRE prep course. Sign up for GRE Question of the Dayemails

Re: If x is an integer, which of the following must be an even i [#permalink]
21 Feb 2017, 16:45

2

This post received KUDOS

Expert's post

Explanation

Again the best strategy is plugging values. Take an odd integer say 1. For simplicity and put the values

A) \(x^2\) — x — 1 = 1 - 1 -1 = -1

B) \(x^2\) — 4x + 6 = 1 -4 + 6 = 3

C) \(x^2\) — 5x + 5 = 1 -5 +5 = 1

D) \(x^2\) + 3x + 8 = 1 + 3 + 8 = 12

E) \(x^2\) + 2x + 10 = Don't need to evaluate this. You already have your answer.

Hence D is correct.

Note: During plugging values if you have found an answer the do not continue plugging in other options. This is very important time saving tactic. Also keep in mind when a question has very generic statements like all integers satisfy "Expression..." It is highlly likely that plugging values is the best strategy.
_________________

Sandy If you found this post useful, please let me know by pressing the Kudos Button

Re: If x is an integer, which of the following must be an even i [#permalink]
12 Aug 2018, 04:26

1

This post received KUDOS

sandy wrote:

Explanation

Again the best strategy is plugging values. Take an odd integer say 1. For simplicity and put the values

A) \(x^2\) — x — 1 = 1 - 1 -1 = -1

B) \(x^2\) — 4x + 6 = 1 -4 + 6 = 3

C) \(x^2\) — 5x + 5 = 1 -5 +5 = 1

D) \(x^2\) + 3x + 8 = 1 + 3 + 8 = 12

E) \(x^2\) + 2x + 10 = Don't need to evaluate this. You already have your answer.

Hence D is correct.

Note: During plugging values if you have found an answer the do not continue plugging in other options. This is very important time saving tactic. Also keep in mind when a question has very generic statements like all integers satisfy "Expression..." It is highlly likely that plugging values is the best strategy.

X can be any integer (i.e. either odd or even) so even though your answer is correct the procedure is incomplete. You would have to check:

1) if D) fits for X as an even integer and than you are finished.

OR

2) if E) fits for x = 1, which it does not [so D) is the only possible option].

Re: If x is an integer, which of the following must be an even i [#permalink]
12 Aug 2018, 20:39

sandy wrote:

Explanation

Again the best strategy is plugging values. Take an odd integer say 1. For simplicity and put the values

A) \(x^2\) — x — 1 = 1 - 1 -1 = -1

B) \(x^2\) — 4x + 6 = 1 -4 + 6 = 3

C) \(x^2\) — 5x + 5 = 1 -5 +5 = 1

D) \(x^2\) + 3x + 8 = 1 + 3 + 8 = 12

E) \(x^2\) + 2x + 10 = Don't need to evaluate this. You already have your answer.

Hence D is correct.

Note: During plugging values if you have found an answer the do not continue plugging in other options. This is very important time saving tactic. Also keep in mind when a question has very generic statements like all integers satisfy "Expression..." It is highlly likely that plugging values is the best strategy.

Re: If x is an integer, which of the following must be an even i [#permalink]
06 Apr 2019, 00:09

fastest way for me was to plug in odd to those X and only one will have even..which is D you can just do it in ur head w/o writing down bunch of E,O stuff which i don't even know what that is.