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# If k is a multiple of 4, which of the following is NOT a possible.....

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Director
Joined: 12 Feb 2015
Posts: 875
If k is a multiple of 4, which of the following is NOT a possible.....  [#permalink]

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15 Aug 2018, 06:42
1
7
00:00

Difficulty:

35% (medium)

Question Stats:

69% (01:42) correct 31% (02:16) wrong based on 121 sessions

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If k is a multiple of 4, which of the following is NOT a possible root of $$x^2 – 6x + k$$ = 0?
A) 1
B) 3−$$\sqrt{5}$$
C) 2
D) 4
E) 3+$$\sqrt{5}$$

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Manish

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Re: If k is a multiple of 4, which of the following is NOT a possible.....  [#permalink]

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15 Aug 2018, 06:48
4
2
Official Solution:-

Step 1: Glance Read Jot
There are several indications that this will be a challenging problem. First, it is a Quadratic Equations problem with an unknown constant. It will be hard to factor in a meaningful way. Second, the question asks which is NOT a possible root. That means there are at least four possible roots to this quadratic, and it’s possible you will have to solve for all of them. Third, some of the answers have square roots in them, meaning the algebra could be complex. On the other hand, the answer choices represent possible values for x, which means that Working Backwards is an option.

Jot down what’s given:
x^2 – 6x + k = 0
k = multiple of 4
Q: x ≠ ?

Step 2: Reflect Organize
Because of the difficulties of this problem, this may be a good question to skip. However, two things strongly suggest that the right solution path is to Work Backwards. First, because it’s asking which answer is NOT a root, four are roots, and you can plug the answer choices into the equation to determine which four work. Second, three of the five answer choices are very simple numbers. Plan to work backwards.

Step 3: Work
Work Backwards
When Working Backwards, it is often best to start with answer choice (B) or (D). But on this problem, identifying that an answer is wrong will not give any clue as to whether the right answer should be larger or smaller. On this problem, you are more likely to save time by starting with any of the three easier answer choices. Save choices (B) and (E) for last.

A) 1 Substitute for x
(1)^2 – 6(1) + k = 0
1 – 6 + k = 0
–5 + k = 0
k = 5

If x is 1, k is not a multiple of 4. Choice (A) is correct. The work for the other choices is shown below in case you started Working Backwards from any other answer choice.

C) 2 Substitute for x
(2)^2 – 6(2) + k = 0
4 – 12 + k = 0
–8 + k = 0
k = 8

Two is a possible root, eliminate choice (C).

D) 4 Substitute for x
(4)^2 – 6(4) + k = 0
16 – 24 + k = 0
–8 + k = 0
k = 8

Four is a possible root, eliminate choice (D).

Don’t test the more challenging answer choices unless you’ve eliminated the easier three, which you should not be able to for this problem. However, the work to test them is shown below.

B) 3 – $$\sqrt{5}$$ Substitute for x
(3 – $$\sqrt{5}$$)2 – 6(3 – $$\sqrt{5}$$) + k = 0
– 4 + k = 0
k = 4

Eliminate choice (B).

E) 3 + $$\sqrt{5}$$ Substitute for x
(3 + $$\sqrt{5}$$)2 – 6(3 + $$\sqrt{5}$$) + k = 0
–4 + k = 0
k = 4

Eliminate choice (E).

The correct answer is (A).
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##### General Discussion
Manager
Joined: 28 Mar 2017
Posts: 57
Location: Sweden
Concentration: Finance, Statistics
Re: If k is a multiple of 4, which of the following is NOT a possible.....  [#permalink]

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18 Aug 2018, 06:42
CAMANISHPARMAR wrote:
Official Solution:-

Step 1: Glance Read Jot
There are several indications that this will be a challenging problem. First, it is a Quadratic Equations problem with an unknown constant. It will be hard to factor in a meaningful way. Second, the question asks which is NOT a possible root. That means there are at least four possible roots to this quadratic, and it’s possible you will have to solve for all of them. Third, some of the answers have square roots in them, meaning the algebra could be complex. On the other hand, the answer choices represent possible values for x, which means that Working Backwards is an option.

Jot down what’s given:
x^2 – 6x + k = 0
k = multiple of 4
Q: x ≠ ?

Step 2: Reflect Organize
Because of the difficulties of this problem, this may be a good question to skip. However, two things strongly suggest that the right solution path is to Work Backwards. First, because it’s asking which answer is NOT a root, four are roots, and you can plug the answer choices into the equation to determine which four work. Second, three of the five answer choices are very simple numbers. Plan to work backwards.

Step 3: Work
Work Backwards
When Working Backwards, it is often best to start with answer choice (B) or (D). But on this problem, identifying that an answer is wrong will not give any clue as to whether the right answer should be larger or smaller. On this problem, you are more likely to save time by starting with any of the three easier answer choices. Save choices (B) and (E) for last.

A) 1 Substitute for x
(1)^2 – 6(1) + k = 0
1 – 6 + k = 0
–5 + k = 0
k = 5

If x is 1, k is not a multiple of 4. Choice (A) is correct. The work for the other choices is shown below in case you started Working Backwards from any other answer choice.

C) 2 Substitute for x
(2)^2 – 6(2) + k = 0
4 – 12 + k = 0
–8 + k = 0
k = 8

Two is a possible root, eliminate choice (C).

D) 4 Substitute for x
(4)^2 – 6(4) + k = 0
16 – 24 + k = 0
–8 + k = 0
k = 8

Four is a possible root, eliminate choice (D).

Don’t test the more challenging answer choices unless you’ve eliminated the easier three, which you should not be able to for this problem. However, the work to test them is shown below.

B) 3 – $$\sqrt{5}$$ Substitute for x
(3 – $$\sqrt{5}$$)2 – 6(3 – $$\sqrt{5}$$) + k = 0
– 4 + k = 0
k = 4

Eliminate choice (B).

E) 3 + $$\sqrt{5}$$ Substitute for x
(3 + $$\sqrt{5}$$)2 – 6(3 + $$\sqrt{5}$$) + k = 0
–4 + k = 0
k = 4

Eliminate choice (E).

The correct answer is (A).

Great approach! Much faster than Competing the square

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If k is a multiple of 4, which of the following is NOT a possible.....  [#permalink]

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18 Aug 2018, 07:08
1
2
CAMANISHPARMAR wrote:
If k is a multiple of 4, which of the following is NOT a possible root of $$x^2 – 6x + k$$ = 0?
A) 1
B) 3−$$\sqrt{5}$$
C) 2
D) 4
E) 3+$$\sqrt{5}$$

Quick observations:-

1) Irrational roots of a QE always occur in pair. Eliminate B & E. (if 3+$$\sqrt{5}$$ is one of the roots of the QE, then 3−$$\sqrt{5}$$ must be the other root of the QE.)

2) Sum of the roots of a quadratic equation=-$$\frac{b}{a}$$=-$$\frac{(-6)}{1}$$=6
& product of the roots=$$\frac{c}{a}$$=$$\frac{k}{1}$$=k, a multiple of 4.
So, the roots could be 2 and 4. (product of roots=2*4=8, a multiple of 4. sum of roots=6=2+4)
Now eliminate C & D.

Ans. (A)

P.S:- It may save 1:30 minutes in real exam.
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PKN

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Intern
Joined: 25 Jul 2012
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Location: India
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GMAT 1: 660 Q44 V37
Re: If k is a multiple of 4, which of the following is NOT a possible.....  [#permalink]

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18 Aug 2018, 07:48
1
For checking whether the given value is a root of the equation, always substitute the given number as value of x and check.
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Posts: 87
GPA: 2.87
Re: If k is a multiple of 4, which of the following is NOT a possible.....  [#permalink]

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18 Aug 2018, 09:05
let's start by plugging in all the answers in the equation and find the value of k which results in non multiple of 4.
only by putting x=1 gives us k=5 which is non multiple of 4.
so Answer is A
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Hasnain Afzal

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Re: If k is a multiple of 4, which of the following is NOT a possible.....   [#permalink] 18 Aug 2018, 09:05
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# If k is a multiple of 4, which of the following is NOT a possible.....

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