Question

If x=√2+1/√2-1,y=√2-1/✓2+1 find the value of x^2+y^2+xy

Answer 1

Heya !!

Identities used :

${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ \\ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \\ \\ (x + y)(x - y) = {x}^{2} - {y}^{2}$

$x = \frac{ \sqrt{2} + 1 }{ \sqrt{2} - 1}$

On rationalizing the denominator we get,

$x = \frac{ \sqrt{2} + 1}{ \sqrt{2} - 1 } \times \frac{ \sqrt{2} + 1}{ \sqrt{2} + 1}$

On rationalizing the denominator we get,

$x = \frac{ {( \sqrt{2} )}^{2} + {(1)}^{2} + 2( \sqrt{2} )(1) }{ {( \sqrt{2} )}^{2} - {(1)}^{2} } \\ \\ x = \frac{2 + 1 + 2 \sqrt{2} }{2 - 1} \\ \\ x = 3 + 2 \sqrt{2}$

On squaring both the sides we get,

${(x)}^{2} = {(3 + 2 \sqrt{2} )}^{2} \\ \\ {x}^{2} = {(3)}^{2} + {(2 \sqrt{2}) }^{2} + 2(3)(2 \sqrt{2} ) \\ \\ {x}^{2} = 9 + 8 + 12 \sqrt{2} \\ \\ {x}^{2} = 17 + 12 \sqrt{2}$

$y = \frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1}$

On rationalizing the denominator we get,

$y = \frac{ \sqrt{2} - 1}{ \sqrt{2} + 1} \times \frac{ \sqrt{2} - 1}{ \sqrt{2} - 1} \\ \\ y = \frac{ {( \sqrt{2} )}^{2} + {(1)}^{2} - 2( \sqrt{2})(1) }{ {( \sqrt{2} )}^{2} - {(1)}^{2} } \\ \\ y = \frac{2 + 1 - 2 \sqrt{2} }{2 - 1} \\ \\ y = 3 - 2 \sqrt{2}$

On squaring both the sides we get,

${(y)}^{2} = {(3 - 2 \sqrt{2}) }^{2} \\ \\ {y}^{2} = {(3)}^{2} + {(2 \sqrt{2}) }^{2} - 2(3)(2 \sqrt{2} ) \\ \\ {y}^{2} = 9 + 8 - 12 \sqrt{2} \\ \\ {y}^{2} = 17 - 12 \sqrt{2}$

Now putting the values

${x}^{2} + {y}^{2} + xy \\ \\ =(17 + 12 \sqrt{2} ) + (17 - 12 \sqrt{2} ) + (3 + 2 \sqrt{2} )(3 - 2 \sqrt{2} ) \\ \\ = 17 + 12 \sqrt{2} + 17 - 12 \sqrt{2} + ( {(3)}^{2} - {(2 \sqrt{2}) }^{2}) \\ \\ = 17 + 17 + (9 - 8) \\ \\ = 34 + (1) \\ \\ = 34 + 1 \\ \\ = 35$

Hope this helps ☺

Answer 2

Answer:

Step-by-step explanation:

X = 2+1+2ROOT2

Y= 3-2ROOT2

X+Y = 6

XY = 9-8=1

[X+Y]^2=X2+Y2 +2XY

36-2 = X2+Y2

34 =X2+Y2

THEN

X2+Y2+XY= 34+1=35