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Posts: 27
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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) \(x^2 + 7x + k\)
GMAT-Club-Forum-rg3ssd0i.png [ 46.66 KiB | Viewed 11114 times ]
(A) \(x^2 + kx - 7\)
(B) \(x^2 - 6x + k\)
(C) \(x^2 + 6x + k\)
(D) \(x^2 - 7x + k\)
(E) \(x^2 + 7x + k\)
Attachment:
GMAT-Club-Forum-rg3ssd0i.png [ 46.66 KiB | Viewed 11114 times ]
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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.
\(x = 3 - \sqrt{2}\)
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.
Posts: 91
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!
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!
General Discussion
GMAT 1: 800 Q51 V51
Posts: 2,161
Updated on: 08 Apr 2025, 13:00
Originally posted by MartyMurray on 09 Oct 2023, 11:47.
Last edited by MartyMurray on 08 Apr 2025, 13:00, edited 5 times in total.
Last edited by MartyMurray on 08 Apr 2025, 13:00, edited 5 times in total.
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
_________________
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
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gmatophobia
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yrozenblum
\(x = 3 - \sqrt{2}\)
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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Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
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
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
Joined: 08 Dec 2023
Last visit: 18 Feb 2026
Posts: 309
Location: India
Concentration: General Management, Human Resources
GPA: 8.88
WE:Engineering (Technology)
Posts: 309
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.
ZIX
Best approach
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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
(x-3)^2=(-rt2)^2
=>x^2-6x+9=2
=x^2-6x+k=0
Option B
Posted from my mobile device
Posts: 1,348
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.
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 \(k=7\).
_________________
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.
yrozenblum
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 \(k=7\).
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yrozenblum
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.
Posts: 5,201
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
(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
yrozenblum
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
Attachment:
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will this approach work for 2 solutions?
ZIX
If we substitute x=3-√2 in the given answer choices
we see that for option B)
x2-6x+k = 0
⇒(3-√2)^2-6(3-√2)+k=0
⇒9+2-6√2-18+6√2+k=0
which gives k=7 which satisfies the condition in question stem (for some integer k)
Correct Answer B
we see that for option B)
x2-6x+k = 0
⇒(3-√2)^2-6(3-√2)+k=0
⇒9+2-6√2-18+6√2+k=0
which gives k=7 which satisfies the condition in question stem (for some integer k)
Correct Answer B
yrozenblum
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