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

Question

If x = 3 - \sqrt{2} , which of the following has the value 0 for some integer k? (A) x^2 + kx - 7 (B) x^2 - 6x + k (C) x^2 + 6x + k (D) x^2 - 7x + k (E) x^2 + 7x + k https://gmatclub.com/forum/download/file.php?id=81453 GMAT-Club-Forum-rg3ssd0i.png

ZIX · Accepted Answer

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.

shwetakoshija · Answer

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

Choice Analysis : Only choice  B  satisfies this!

Hope this helps!

MartyMurray · Answer

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

Hacker Approach

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  x^2 , let's calculate  x^2 .

x = 3 - \sqrt{2}

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

Now, since  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} .

Let's scan the answer choices again.

A)  x^2 + kx - 7

B)  x^2 - 6x + k

C)  x^2 + 6x + k

D)  x^2 - 7x + k

E)  x^2 + 7x + k

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

Correct Answer:   B

gmatophobia · Answer

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

Jahnavi_M · 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

Posted from my mobile device

samarpan.g28 · Answer

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.

saynchalk · Answer

Best approach

Oppenheimer1945 · Answer

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

GMATGuruNY · Answer

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 .

B

btsaami · Answer

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.

GmatKnightTutor · Answer

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

(A)  x^2 + kx - 7

(B)  x^2 - 6x + k

(C)  x^2 + 6x + k

(D)  x^2 - 7x + k

(E)  x^2 + 7x + k

Interesting question. In a way, it basically comes down to trying to get rid of the  \sqrt{2}  variable.

One way to do this question is by first expanding and figuring out what  x^2  is

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

Now we need to look for an answer choice that AFTER THE  x^2  part has something that will get rid of  -6 \sqrt{2} . We're not really concerned with the integer values because we can use any value of k to "get rid" of their sum.

We basically need a  +6 \sqrt{2}  to counter the  -6 \sqrt{2}

Only (B) does this ->  11 - 6\sqrt{2}   -18 + 6 \sqrt{2} + k

-7 + k

Therefore, (B) will become 0 when we let k = 7

GMATinsight · Answer

CONCEPT: if x = 3 - \sqrt{2} is one of the roots of the equation then other root must be CONJUGATE of it i,e, other root much be x = 3 + \sqrt{2} i.e. Equation should be written as (x - first root)*(x- second root) = 0 i.e. (x - 3 + √2)*(x - 3 - √2) = 0 i.e. x^2 - 6x + Constant = 0 i.e. Only Option B matches with the coefficient of x hence Answer: Option B GMAT-Club-Forum-zgudvrwd.png

esseeveniet · Answer

will this approach work for 2 solutions?

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

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

Prep Toolkit

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