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605-655 (Medium)|   Algebra|   Roots|      
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BrentGMATPrepNow
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If x = 1/(√11 + √10) and y = 01 Mar 2017, 09:08
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45% (medium)

Question Stats:

74% (02:30) correct 26% (02:53) wrong based on 477 sessions

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If \(x = \frac{1}{√11 + √10}\) and \(y = \frac{1}{√11 - √10}\), then what is the value of x2 - xy + y2?

A) 39
B) 41
C) 43
D) 45
E) 47

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If x = 1/(√11 + √10) and y = 01 Mar 2017, 09:24
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GMATPrepNow
If \(x = \frac{1}{√11 + √10}\) and \(y = \frac{1}{√11 - √10}\), then what is the value of x² - xy + y²?

A) 39
B) 41
C) 43
D) 45
E) 47

*kudos for all correct solutions
Show more

Hi

Can't paste a fraction with so many square roots :cry:

\(x = \frac{1}{\sqrt{11} + \sqrt{10}} * (\sqrt{11} - \sqrt{10})/(\sqrt{11} - \sqrt{10}) = \sqrt{11} - \sqrt{10}\)

\(y = \frac{1}{\sqrt{11} - \sqrt{10}} * (\sqrt{11} + \sqrt{10})/(\sqrt{11} + \sqrt{10}) = \sqrt{11} + \sqrt{10}\)

\(x^2 - xy + y^2 = (\sqrt{11} - \sqrt{10})^2 - (\sqrt{11} - \sqrt{10})(\sqrt{11} + \sqrt{10}) + (\sqrt{11} + \sqrt{10})^2 =\)
\(11 - 2*\sqrt{11}*\sqrt{10} + 10 - (11 - 10) + 11 + 2*\sqrt{11}*\sqrt{10} + 10 = 41\)

Answer B
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sravankumar0911
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If x = 1/(√11 + √10) and y = 01 Mar 2017, 09:58
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x^2-xy+y^2 can be written as (x-y)^2 + xy....plug in the values the solution will simplify correct answer 41


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If x = 1/(√11 + √10) and y = 01 Mar 2017, 10:46
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If \(x = \frac{1}{√11 + √10}\) and \(y = \frac{1}{√11 - √10}\), then what is the value of x² - xy + y²?

A) 39
B) 41
C) 43
D) 45
E) 47

*kudos for all correct solutions
Show more

\(x^2 = (\frac{1}{√11 + √10})^2\)

Or, \(x^2 = \frac{1}{11 + 10 + 2√110}\) \(= \frac{1}{21 + 2√110}\)

\(y^2 = (\frac{1}{√11 - √10})^2\)

Or, \(y^2 = \frac{1}{11 + 10 - 2√110}\) \(= \frac{1}{21 - 2√110}\)

\(xy = \frac{1}{√11 + √10}*\frac{1}{√11 - √10}\)

So, \(xy = \frac{1}{11 - 10}\) \(= 1\)


Now, \(x^2 - xy + y^2 = \frac{1}{21 + 2√110} - 1 + \frac{1}{21 - 2√110} = 41\)

Thus, answer must be (B) 41
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If x = 1/(√11 + √10) and y = 04 Oct 2025, 03:36
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Most efficient way is to analyse the Question and see if there's a way to simplify.

Q can be re-written as \(x^2\) + \(y^2\) - xy - xy + xy ==> \((x-y)^2\) + xy

When you solve for (x-y) and (xy) you will notice that the denominator can be reduced to \(a^2 - b^2\) form

Option B. 41.
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