Question

If x = √2+ 1/ √2−1 and y =√2− 1 /√2+ 1 . Find the value of x^2 + y^2 + xy

Answer 1

=(root2+1/root2-1)^2+(root2-1/root2+1)^2+(root2+1/root2-1)×(root2-1/root2+1)
=(2+1+2root2/2+1-2root2)+(2+1-2root2/2+1+2root2)
=3+2root2+3-2root2/9-8
=9/1=9
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Answer 2

Answer:

⇒ x² + y² + xy = 35

Step-by-step explanation:

Given ,

$= > x = \frac{ \sqrt{2} + 1 }{ \sqrt{2} - 1} \: and \: y = \frac{ \sqrt{2} - 1}{ \sqrt{2} + 1}$

To Find :-

⇒x² + y² + xy

Solution :-

Rationalising 'x' :-

$= > \frac{ \sqrt{2} + 1 }{ \sqrt{2} - 1} \times \frac{( \sqrt{2} + 1)}{(\sqrt{2} + 1)}$

$= > \frac{( \sqrt{2} + 1)^{2} }{( \sqrt{2})^{2} - (1)^{2} }$

$= > \frac{( \sqrt{2})^{2} + (1)^{2} + 2 \times \sqrt{2} }{2 - 1}$

$= > 2 + 1 + 2 \sqrt{2}$

$= > 3 + 2 \sqrt{2}$

Rationalising 'y' :-

$= > \frac{ \sqrt{2} - 1}{ \sqrt{2} + 1} \times \frac{( \sqrt{2} - 1)}{( \sqrt{2} - 1) }$

$= > \frac{( \sqrt{2} - 1)^{2} }{( \sqrt{2})^{2} - (1)^{2} }$

$= > \frac{( \sqrt{2})^{2} + (1) ^{2} - 2 \times \sqrt{2} }{2 - 1}$

$= > 2 + 1 - 2 \sqrt{2}$

$= > 3 - 2 \sqrt{2}$

Hence ,

⇒ x² + y² + xy = ( x + y )² - xy

⇒[( 3 + √2) + ( 3 - √2 )]² - ( 3+ 2√2) ( 3 - 2√2 )

⇒(6)² - [(3)² -(2√2)²]

⇒36 - ( 9 - 8 )

⇒36 - 1

⇒ 35