Question

If x=root 2+1/root2-1 and y=root2-1/root2+1 then find the value of x square+y square+xy

Answer 1

Answer:for x, rationalise the denominator then we get 3+2root2

And for y rationalise the denominator then we get 3-2root2

Now, put the values of x and y in the polynomial, after the calculating we get 35. The answer is 35.

Step-by-step explanation:

Answer 2

The value of x² + y² + xy will be equal to 35.

Given:

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

To Find:

The value of x² + y² + xy

Solution:

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

→ We would put the values of 'x' and 'y' in the expression to get the value of x² + y² + xy:

Formula to be used:

• (a+b)(a-b) = a² - b²

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

                                                $=\frac{2+1+2\sqrt{2} }{2+1-2\sqrt{2} } +\frac{2+1-2\sqrt{2} }{2+1+2\sqrt{2} }+(\frac{\sqrt{2} +1}{\sqrt{2} -1} )(\frac{\sqrt{2} -1}{\sqrt{2}+1 } ) \\\\=\frac{(2+1+2\sqrt{2})^2+(2+1-2\sqrt{2})^2 }{(2+1-2\sqrt{2})(2+1+2\sqrt{2} ) } +1\\\\=\frac{(3+2\sqrt{2})^2+(3-2\sqrt{2})^2 }{(3-2\sqrt{2})(3+2\sqrt{2} ) } +1\\\\=\frac{(9+8+12\sqrt{2})+ (9+8-12\sqrt{2} )}{3^{2} -(2\sqrt{2} )^{2} } +1\\\\=\frac{9+8+9+8}{9-8} +1\\\\=34+1\\\\=35$

Hence the value of x² + y² + xy comes out to equal to 35.

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