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:

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

Posted from my mobile device

For a quadratic to yield a value of 0, it must have a factor equal to 0.

If x=5 and x=2 each yield a value of 0 for a certain quadratic, then one factor of the quadratic could be (x-5) and the other (x-2):
(x-5) --> x=5 makes the factor equal to 0 --> (5-5) = 0
(x-2) --> x=2 makes the factor equal to 0 --> (2-2) = 0
Resulting quadratic:
\((x-5)(x-2) = x^2-7x+10\)

Implication:
If x=k yields a value of 0 for a quadratic, then one factor of the quadractic could be (x-k) such that x-k=0.
In accordance with the principle above, one factor of the quadratic could be \((x - (3 - \sqrt{2}))\) such that \((x - (3 - \sqrt{2}))=0\).

Thus:
\((x - (3 - \sqrt{2}))=0\)
\(x - 3 + \sqrt{2}=0\)
\(x - 3 = -\sqrt{2}\)
\((x - 3)^2 = ( -\sqrt{2})^2\)
\(x^2 - 6x + 9 = 2\)
\(x^2 - 6x + 7 = 0\)

The resulting quadratic is a match for \(x^2-6x+k\), where \(k=7\).

­

Here's a really quick and efficient solution -

Observation 1: All choices are quadratic expressions.
Recall Concept 1: For a quadratic equation, Irrational roots always occur in conjugate pairs.

So, If 3 - √2 is a root, then so is 3 + √2.

Recall Concept 2: Sum of roots of a quadratic equation = -(Coefficient of x)/(Coefficient of x^2)

So, (3 + √2) + (3 - √2) = 6 = -b/a [Say]

Choice Analysis: Only choice B satisfies this!

Hope this helps!

Hello @Bunuel @KarishmaB @MartyMurray
Can i use below concept in this sum? Is below method correct?
­If one root of a quadratic equation is irrational (or a surd), then the other root is always irrational and is the conjugate of the first root

a= 1 as seen from the options.

We will have to look at option 1 and remaining options.

Sum of roots= -b/a
3-\sqrt{2} + 3+ \sqrt{2}= -b
Hence, b=-6
Substituting in quadratic equation of the form x^2 + bx +c, we get x^2 - 6x+ k.

To verify if option 1 is also correct or not, i can multiple both the roots {3-\sqrt{2} } x {3+ \sqrt{2}} = 9-2 =7 c= 7 hence 2 is the answer.­