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Senior | Next Rank: 100 Posts
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by aarati » Mon Aug 09, 2010 8:45 pm
If x=\frac{2}{3+\sqrt{7}}, then (x-3)^2 is equal to:

(a) 1
(b)3
(c)6
(d)7

help me to solve it.......
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by 4GMAT_Mumbai » Mon Aug 09, 2010 8:55 pm
Hi,

Ignorance disclaimers ...

What is "\frac"? What is "\sqrt"?
Naveenan Ramachandran
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
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by aarati » Mon Aug 09, 2010 9:16 pm
4GMAT_Mumbai wrote:Hi,

Ignorance disclaimers ...

What is "\frac"? What is "\sqrt"?
If x=2/(3+sqrt{7}), then (x-3)^2 is equal to:

(a) 1
(b)3
(c)6
(d)7
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by 4GMAT_Mumbai » Mon Aug 09, 2010 9:23 pm
Hi,

Unlikely to be GMAT level ...

x - 3 = (2/(3+sqrt{7})) - 3

= (-7-3sqrt(7))/(3+sqrt(7))

Squaring the previous term will lead to 7.
Naveenan Ramachandran
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
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by aarati » Tue Aug 10, 2010 11:52 pm
4GMAT_Mumbai wrote:Hi,

Unlikely to be GMAT level ...

x - 3 = (2/(3+sqrt{7})) - 3

= (-7-3sqrt(7))/(3+sqrt(7))

Squaring the previous term will lead to 7.


thank u very much... can u explain me in step by step procedure please....
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by zareentaj » Wed Aug 11, 2010 5:42 am
4GMAT_Mumbai wrote:Hi,

Unlikely to be GMAT level ...

x - 3 = (2/(3+sqrt{7})) - 3

= (-7-3sqrt(7))/(3+sqrt(7))

Squaring the previous term will lead to 7.
Thanks.
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by kmittal82 » Wed Aug 11, 2010 7:07 am
aarati wrote:
4GMAT_Mumbai wrote:Hi,

Unlikely to be GMAT level ...

x - 3 = (2/(3+sqrt{7})) - 3

= (-7-3sqrt(7))/(3+sqrt(7))

Squaring the previous term will lead to 7.


thank u very much... can u explain me in step by step procedure please....
In problems like these, you have to rationalize the denominator i.e. make it into an integer if you can (look here https://www.math.unt.edu/mathlab/emathla ... action.htm)

so, for this example, multiply and divide by 3 - sqrt(7)

x = 2/(3+sqrt(7)) * (3-sqrt(7))/(3-sqrt(7))

The denominator now takes the form a^2 - b^2, and in this case becomes 9 -7 = 2

x = 2*((3 - sqrt(7))/2

x - 3 = sqrt(7)

(x-3)^2 = 7

I think this sort of question could easily be in the GMAT, in the 500-600 level (I have seen a few in some practice tests which rely on denominator rationalisation)
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by aarati » Wed Aug 11, 2010 10:03 pm
kmittal82 wrote:
aarati wrote:
4GMAT_Mumbai wrote:Hi,

Unlikely to be GMAT level ...

x - 3 = (2/(3+sqrt{7})) - 3

= (-7-3sqrt(7))/(3+sqrt(7))

Squaring the previous term will lead to 7.


thank u very much... can u explain me in step by step procedure please....
In problems like these, you have to rationalize the denominator i.e. make it into an integer if you can (look here https://www.math.unt.edu/mathlab/emathla ... action.htm)

so, for this example, multiply and divide by 3 - sqrt(7)

x = 2/(3+sqrt(7)) * (3-sqrt(7))/(3-sqrt(7))

The denominator now takes the form a^2 - b^2, and in this case becomes 9 -7 = 2

x = 2*((3 - sqrt(7))/2

x - 3 = sqrt(7)

(x-3)^2 = 7

I think this sort of question could easily be in the GMAT, in the 500-600 level (I have seen a few in some practice tests which rely on denominator rationalisation)
thank u very much....this the only way to solve this type of problems....
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