Official Solution:-
Step 1: Glance Read JotThere 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 OrganizeBecause 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: WorkWork 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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