Question

if x=root3-1/root3+1,y=3+2root2/3-2root2 then find the value of x+y

Answer 1

Using the identities :
${(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ \\ (a + b)(a - b) = {a}^{2} - {b}^{2}$

Now,

$x = \frac{ \sqrt{3} - 1}{ \sqrt{3} + 1 }$

On rationalizing the denominator we get,

$x = \frac{ \sqrt{3} - 1}{ \sqrt{3} + 1 } \times \frac{ \sqrt{3} - 1 }{ \sqrt{3} - 1 } \\ \\ x = \frac{ {( \sqrt{3} )}^{2} + {(1)}^{2} - 2( \sqrt{3} )(1)}{ {( \sqrt{3} )}^{2} - {(1)}^{2} } \\ \\ x = \frac{3 + 1 - 2 \sqrt{3} }{3 - 1} \\ \\ x = \frac{4 - 2 \sqrt{3} }{2} \\ \\ x = 2 - \sqrt{3}$

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

On rationalizing the denominator we get,

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

Now,

$(2 - \sqrt{3} ) + (17 + 12 \sqrt{2} ) \\ \\ = 2 - \sqrt{3} + 17 + 12 \sqrt{2} \\ \\ = - \sqrt{3} + 12 \sqrt{2} + 19$