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

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: