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\[\text{Let } x = \frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} - \sqrt{5}}, \text{ and } y = \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}}\]

What is \((x^2 - xy + y^2)\) equal to ?

This question was previously asked in
CDS 1 2026 Maths Question Paper (12-Apr-2026)
The correct answer is

481

We are given the expressions for \(x\) and \(y\):

  • \(x = \frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} - \sqrt{5}}\)
  • \(y = \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}}\)

We need to find the value of \(x^2 - xy + y^2\).

Simplify Expressions for x and y

To simplify \(x\), we rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is \(\sqrt{6} + \sqrt{5}\):

\(x = \frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} - \sqrt{5}} \times \frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} + \sqrt{5}}\)
\(x = \frac{(\sqrt{6} + \sqrt{5})^2}{(\sqrt{6})^2 - (\sqrt{5})^2}\)
\(x = \frac{6 + 2(\sqrt{6})(\sqrt{5}) + 5}{6 - 5}\)
\(x = \frac{11 + 2\sqrt{30}}{1}\)
\(x = 11 + 2\sqrt{30}\)

Similarly, to simplify \(y\), we rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is \(\sqrt{6} - \sqrt{5}\):

\(y = \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}} \times \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} - \sqrt{5}}\)
\(y = \frac{(\sqrt{6} - \sqrt{5})^2}{(\sqrt{6})^2 - (\sqrt{5})^2}\)
\(y = \frac{6 - 2(\sqrt{6})(\sqrt{5}) + 5}{6 - 5}\)
\(y = \frac{11 - 2\sqrt{30}}{1}\)
\(y = 11 - 2\sqrt{30}\)

Calculate Intermediate Values

We can express \(x^2 - xy + y^2\) using \(x+y\) and \(xy\). Note that \(x^2 - xy + y^2 = (x^2 + 2xy + y^2) - 3xy = (x+y)^2 - 3xy\).

First, calculate \(x+y\):

\(x + y = (11 + 2\sqrt{30}) + (11 - 2\sqrt{30})\)
\(x + y = 11 + 11 + 2\sqrt{30} - 2\sqrt{30}\)
\(x + y = 22\)

Next, calculate \(xy\):

\(xy = (11 + 2\sqrt{30})(11 - 2\sqrt{30})\)

This is in the form \((a+b)(a-b) = a^2 - b^2\).

\(xy = (11)^2 - (2\sqrt{30})^2\)
\(xy = 121 - (4 \times 30)\)
\(xy = 121 - 120\)
\(xy = 1\)

Calculate Final Expression Value

Now substitute the values of \(x+y\) and \(xy\) into the expression \((x+y)^2 - 3xy\):

\(x^2 - xy + y^2 = (x+y)^2 - 3xy\)
\(= (22)^2 - 3(1)\)
\(= 484 - 3\)
\(= 481\)

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Important Questions from Algebra

  1. The difference between the two positive numbers x and y where x > y, is 25% of x. If the value of y is 15, then the value of x is:

  2. If p2 + q2 - r2 = 0, then the value of (p6 + q6 - r6) ÷ p2q2r2 is:

  3. If √2 + √x = √3, then the value of x is equal to:

  4. The sum of two numbers is 20 and their difference is 2.5. Ratio of these numbers will be:

  5. If \(\rm \frac{\sqrt{19 - x \sqrt{12}}}{1} = \sqrt 4 - \sqrt 3\)  then the value of x is equal to:

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