If x = 3 - 2^(1/2), which of the following has the value 0 for some

Here is an easier approach:­

\(x = 3 - \sqrt{2}\)

Bring 3 to the left side and square both the sides. 

\((x-3)^2 = (-\sqrt{2})^2\)

solve and you get:

\(x^2 + 9 - 6x = 2\)
\(x^2 - 6x +7 = 0\)

Only option B matches. ­­­­

If \(x = 3 - \sqrt{2}\), which of the following has the value 0 for some integer k?

Scanning the answer choices, we see that, for a choice to have the value 0, the first two terms of the choice must have an integer value so that the first two terms - 7 or + integer k can equal 0.

Since all the choices include x², let's calculate x².

\(x = 3 - \sqrt{2}\)

\(x^2 = 9 - 6\sqrt{2} + 2 = 11 - 6\sqrt{2}\).

A) x² + kx - 7

This equals the following:

\(11 - 6\sqrt{2} + 3k - k\sqrt{2} -7\)

If \(k = -6\), we can eliminate \(\sqrt{2}\). So, in that case, we will have an integer value, but it is \(11 - 18 - 7 = -14\).

So, we can't get \(0\).

B) x² - 6x + k

In this case, the first two terms equal the following:

\(11 - 6\sqrt{2} - 18 + 6\sqrt{2} = -7\)

So, if \(k = 7, x^2 - 6x + k = 0\)

Thus, for "some integer," \(7\), this choice has a value of \(0\).

C) x² + 6x + k

In this case, the first two terms equal the following:

\(11 - 6\sqrt{2} + 18 - 6\sqrt{2} = 29 - 12\sqrt{2}\)

Since \(29 - 12\sqrt{2}\) is not an integer value, there is no integer k we can add to it to get a value of \(0\).

D) x² - 7x + k

In this case, the first two terms equal the following:

\(11 - 6\sqrt{2} - 21 + 7\sqrt{2} = -10 + \sqrt{2}\)

Since \(-10 + \sqrt{2}\) is not an integer value, there is no integer k we can add to it to get a value of \(0\).

E) x² + 7x + k

In this case, the first two terms equal the following:

\(11 - 6\sqrt{2} + 21 - 7\sqrt{2} = 32 - 13\sqrt{2}\)

Since \(32 - 13\sqrt{2}\) is not an integer value, there is no integer k we can add to it to get a value of \(0\).

Now, having seen how all the choices work, it's good to also be aware that we don't have to go through all the choices to get the answer to this question. Rather, once we see that \(x^2 = 11 - 6\sqrt{2}\), we can tell that, to get a value of \(0\) by adding or subtracting an integer, we need to eliminate the \(- 6\sqrt{2}\).

The only choice that can do that is choice (B), which includes \(-6x\), which includes \(6\sqrt{2}\). So, the only choice that can work is (B).

Correct Answer:
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\(x = 3 - \sqrt{2}\)

Squaring both sides we get

\(x^2 = (3 - \sqrt{2})^2\)

\(x^2 = 9 + 2 - 6\sqrt{2}\)

\(x^2 = 11 - 6\sqrt{2}\)

We know that \(k\) is an integer. Therefore, we can choose a value of k so that the integer part of the equation cancels out and results in a value of 0. However, the equation contains a non-integer portion, \(- 6\sqrt{2}\). To cancel \(- 6\sqrt{2}\), we need a value of \(+ 6\sqrt{2}\). As only x contains the value \(-\sqrt{2}\), we need to multiply x by -6 so as to get the value \(+6\sqrt{2}\).

Among the given options only Option B contains \(-6x\).

Option B
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I was thinking in terms of (-b + sqrt(b^2 - 4ac))/2a and (-b - sqrt(b^2 - 4ac))/2a

One of the x = 3 - sqrt(2)
Since, all the options have a = 1 in ax^2 + bx + c

=> x = (6 - sqrt(8))/2
=> x = (6 - sqrt(36 - 4 * 7))/2

=> b = -6, C = 7

So, x ^ 2 - 6 x + 7 = 0

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This is how I found the solution. I calculated x^2 at first which yielded 11-6√ 2. From the options, I calculated 6x=18-6√ 2. Now (11-6√ 2)-(18-6√ 2)=-7+k. I checked all options, B seemed a close match. If k=7 then -7+7=0. Therefore, B is the correct option.

 

­Best approach

Get rid of irrational terms:

(x-3)^2=(-rt2)^2
=>x^2-6x+9=2
=x^2-6x+k=0

Option B

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