Question

if x=√3-√2/√3+√2 and y=√3+√2/√3-√2 then x²+xy+y²

explain clearly​

Answer 1

Answer:

99

Step-by-step explanation:

As per the provided information in the given question, we have :

• $\sf{x=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}$

• $\sf{y=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}$

We have been asked to calculate the value of x² + xy + y².

In order to tackle this question, firstly we'll rationalise the denominator of the given fraction of the values of x and y in the question.

Rationalisation is the process of making the denominator rational. It is done by multiplying the rationalising factor of the denominator with both the numerator and the denominator. Rationalising factor is the term which is when multiplied with the numerator and the denominator, the denominator becomes rational.

Coming back to the question, let's rationalise the denominator of the values of x and y!

Rationalising the denominator of the value of x :

$\implies\sf{x=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}$

Here, the denominator is (√3 + 2) which is in the form of (a + b). The rationalising factor of (a + b) is (a ― b). So, the rationalising factor of (√3 + 2) is (√3 ― 2).

Multiplying (√3 ― 2) with both the numerator and the denominator.

$\implies\sf{x=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}}$

Rearranging the terms. We can write it as,

$\implies\sf{x=\dfrac{(\sqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}}$

Or,

$\implies\sf{x=\dfrac{(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}}$

We know that,

• (a — b)² = a² + b² — 2ab
• (a + b)(a — b) = a² — b²

$\implies\sf{x=\dfrac{(\sqrt{3})^2+(\sqrt{2})^2-2(\sqrt{3}\times \sqrt{2})}{(\sqrt{3})^2-(\sqrt{2})^2}}$

Writing the squares of the numbers in the numerator and the denominator. Also, performing multiplication in the numerator.

$\implies\sf{x=\dfrac{3+2-2\sqrt{6}}{3-2}}$

Performing addition in numerator and subtraction in denominator.

$\implies\boxed{\sf{x=5-2\sqrt{6}}}\dots [1]$

Rationalising the denominator of the value of y :

$\implies\sf{y=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}$

Here, the denominator is (√3 ― 2) which is in the form of (a ― b). The rationalising factor of (a ― b) is (a + b). So, the rationalising factor of (√3 ― 2) is (√3 +2).

Multiplying (√3 + 2) with both the numerator and the denominator.

$\implies\sf{y=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}}$

Rearranging the terms. We can write it as,

$\implies\sf{y=\dfrac{(\sqrt{3}+\sqrt{2})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}}$

Or,

$\implies\sf{y=\dfrac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}}$

We know that,

• (a + b)² = a² + b² + 2ab
• (a + b)(a — b) = a² — b²

$\implies\sf{x=\dfrac{(\sqrt{3})^2+(\sqrt{2})^2+2(\sqrt{3}\times \sqrt{2})}{(\sqrt{3})^2-(\sqrt{2})^2}}$

Writing the squares of the numbers in the numerator and the denominator. Also, performing multiplication in the numerator.

$\implies\sf{y=\dfrac{3+2+2\sqrt{6}}{3-2}}$

Performing addition in numerator and subtraction in denominator.

$\implies\boxed{\sf{y=5+2\sqrt{6}}}\dots [2]$

Finding the value of x² + xy + y² :

$\implies\sf{x^2 + xy + y^2}$

Substitute the values of x and y.

$\implies\sf{(5-2\sqrt{6})^2 + \Big \{ (5-2\sqrt{6})(5+2\sqrt{6})\Big \}+ (5+2\sqrt{6})^2}$

We know that,

• (a + b)² = a² + b² + 2ab
• (a ― b)² = a² + b² ― 2ab
• (a + b)(a — b) = a² — b²

Hence,

Value of x² :

$\implies\sf{x^2 = (5-2\sqrt{6})^2 }$

$\implies\sf{x^2 = (5)^2 + (2\sqrt{6})^2 - 2(5\times 2\sqrt{6}) }$

$\implies\sf{x^2 = 25 + 24 - 2(10\sqrt{6}) }$

$\implies\underline{\sf{\green{x^2 = 49 - 20\sqrt{6} }}}$

Value of xy :

$\implies\sf{xy = (5-2\sqrt{6})(5+2\sqrt{6}) }$

$\implies\sf{xy = (5)^2-(2\sqrt{6})^2 }$

$\implies\sf{xy = 25-24 }$

$\implies\underline{\sf{\green{xy = 1 }}}$

Value of y²

$\implies\sf{y^2 = (5+2\sqrt{6})^2 }$

$\implies\sf{y^2 = (5)^2 + (2\sqrt{6})^2 + 2(5\times 2\sqrt{6}) }$

$\implies\sf{y^2 = 25 + 24 + 2(10\sqrt{6}) }$

$\implies\underline{\sf{\green{y^2 = 49 + 20\sqrt{6} }}}$

Now,

$\implies\sf{x^2 + xy + y^2}$

$\implies\sf{(49 - 20\sqrt{6})+ 1 + (49 +20\sqrt{6})}$

$\implies\sf{49 - 20\sqrt{6}+ 1 + 49 +20\sqrt{6}}$

$\implies\sf{49 + 1 + 49 }$

$\implies\underline{\boxed{\sf{\red{99}}}}$

Therefore, the value of x² + xy + y² here is 99.

Answer 2

As per the given information :-

$\sf \bull \: x = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

$\sf \bull \: y = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

To find the value of :-

$\sf \bull \: {x}^{2} + xy + {y}^{2}$

Rationalising the denominator of x :-

$\sf \bull \: \frac{ \sqrt{ 3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{ 3} - \sqrt{2} } \\ \\ \sf \bullet \: \frac{( \sqrt{ 3} - \sqrt{2} ) ^{2} }{( \sqrt{3} + \sqrt{2} )( \sqrt{3} - \sqrt{2}) }$

$\sf \bull \: \dfrac{ { \sqrt{3} }^{2} - 2( \sqrt{3} \times \sqrt{2}) + { \sqrt{2} }^{2} }{ { \sqrt{3} }^{2} - \sqrt{2} ^{2} } \\ \\ \sf \bull \: \dfrac{3 - 2 \times \sqrt{6} + 2}{3 - 2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \bull \: \frac{5 - 2 \sqrt{6} }{1} = 5 - 2 \sqrt{6} \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Rationalising the denominator of y

$\sf \bull \: \frac{ \sqrt{ 3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{ 3} + \sqrt{2} } \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \bullet \: \frac{( \sqrt{ 3} + \sqrt{2} ) ^{2} }{( \sqrt{3} - \sqrt{2} )( \sqrt{3} + \sqrt{2}) } \: \: \: \: \: \: \: \: \\ \\ \sf \bull \: \dfrac{ { \sqrt{3} }^{2} + 2( \sqrt{3} \times \sqrt{2}) + { \sqrt{2} }^{2} }{ { \sqrt{3} }^{2} - \sqrt{2} ^{2} } \\ \\ \sf \bull \: \dfrac{3 + 2 \times \sqrt{6} + 2}{3 - 2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \bull \: \frac{5 + 2 \sqrt{6} }{1} = 5 + 2 \sqrt{6} \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Therefore the value of x and y is $5 - 2 \sqrt{6}$ and $5 + 2 \sqrt{6}$ respectively.

Now we will put the obtained value in the equation x²+ xy+y² and equate it to determine the value .

$\sf : \implies \: {x}^{2} + xy + {y}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf : \implies \: (5 - 2 \sqrt{6} ) ^{2} + (5 + 2 \sqrt{6} )(5 - 2 \sqrt{6} ) + (5 + 2 \sqrt{6} ) ^{2}$

• We will calculate the value of x²

$\sf : \implies \:(5 - 2 \sqrt{6} ) ^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf : \implies \: {(5)}^{2} - (2 \times 5 \times 2 \sqrt{6} ) + {(2 \sqrt{6} )}^{2} \\ \\ \sf : \implies \:25 - 2(10 \sqrt{6} ) + 4 \times 6 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf : \implies \: 25 + 24 - 20 \sqrt{6} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf : \implies \: 49 - 20 \sqrt{6} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

• We will calculate the value of xy

$\sf : \implies \: (5 - 2 \sqrt{6} )(5 + 2 \sqrt{6} ) \\ \\ \sf : \implies \: {(5)}^{2} - {(2 \sqrt{6}) }^{2} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf : \implies \: 25 - 24 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf : \implies \: 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

• Now we will find the value of y²

$\sf : \implies \: {(5 + 2 \sqrt{6} )}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf : \implies \: {(5)}^{2} + 2 \times (5 \times 2 \sqrt{6} ) + {(2 \sqrt{6} )}^{2} \\ \\ \sf : \implies \: 25 + 2 \times (10 \sqrt{6} ) + 4 \times 6 \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf : \implies \: 25 + 24 + 20 \sqrt{6} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf : \implies \: 49 + 20 \sqrt{6} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Now we will substitute these value and obtain our required answer :-

$\sf : \implies \: (49 - 20 \sqrt{6} ) + 1 + (49 + 20 \sqrt{6} ) \\ \\ \sf : \implies \: (49 \: \: \cancel{ - 20 \sqrt{6}} ) + 1 + (49 \: \cancel{+ 20 \sqrt{6}} ) \: \\ \\\sf : \implies \:49 + 49 + 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf : \implies \: \bigstar \: \underline{\boxed{\red{99 }}} \: \bigstar \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Identity Used :-

$\bigstar \boxed {\begin {array}{cc} \sf 1. \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ \\ \sf 2. \: (x - y) ^{2} = {x}^{2} - 2xy + {y}^{2} \\ \\ \sf 3. \: (x - y)(x + y) = {x}^{2} - {y}^{2} \end{array}} \bigstar$