Question

If x = √3-√2/√3+√2 and y = √3+√2/√3-√2, then find the value of x square + y square + xy.​

Answer 1

`\large\underline{\sf{Solution-}}`

(i)↓

`\sf x=(\sqrt3-\sqrt2)/(\sqrt3+\sqrt2)`

On rationalising,

`\sf\longmapsto x=(\sqrt3-\sqrt2)/(\sqrt3+\sqrt2)*(\sqrt3-\sqrt2)/(\sqrt3-\sqrt2)`

So,

`\sf\longmapsto x=((\sqrt3-\sqrt2)^2)/((\sqrt3+\sqrt2)(\sqrt3-\sqrt2))`

We know that,

• 
  `\sf (a-b)(a+b)=a^2-b^2`

And,

• 
  `\sf (a-b)^2=a^2-2ab+b^2`

So,

`\sf\longmapsto x=((\sqrt3)^2-2(\sqrt3)(\sqrt2)+(\sqrt2)^2)/((\sqrt3)^2-(\sqrt2)^2)`

`\sf\longmapsto x=(3-2\sqrt6+2)/(3-2)`

`\sf\longmapsto x=(5-2\sqrt6)/(1)`

Hence,

`\sf\longmapsto x=5-2\sqrt6`

(ii)↓

`\sf y=(\sqrt3+\sqrt2)/(\sqrt3-\sqrt2)`

On rationalising,

`\sf\longmapsto y=(\sqrt3+\sqrt2)/(\sqrt3-\sqrt2)*(\sqrt3+\sqrt2)/(\sqrt3+\sqrt2)`

So,

`\sf\longmapsto y=((\sqrt3+\sqrt2)^2)/((\sqrt3+\sqrt2)(\sqrt3-\sqrt2))`

We know that,

• 
  `\sf (a-b)(a+b)=a^2-b^2`

And,

• 
  `\sf (a+b)^2=a^2+2ab+b^2`

So,

`\sf\longmapsto y=((\sqrt3)^2+2(\sqrt3)(\sqrt2)+(\sqrt2)^2)/((\sqrt3)^2-(\sqrt2)^2)`

`\sf\longmapsto y=(3+2\sqrt6+2)/(3-2)`

`\sf\longmapsto y=(5+2\sqrt6)/(1)`

Hence,

`\sf\longmapsto y=5+2\sqrt6`

Now, finding the value of,↓↓

`\sf x^2+y^2+xy`

Now, substituting the values of x and y in equation,

`\sf\longmapsto (5-2\sqrt6)^2+(5+2\sqrt6)^2+(5-2\sqrt6)(5+2\sqrt6)`

So,

`\sf\longmapsto [(5)^2-2(5)(2\sqrt6)+(2\sqrt6)^2]+[(5)^2+2(5)(2\sqrt6)+(2\sqrt6)^2]+[(5)^2-(2\sqrt6)^2]`

`\sf\longmapsto 25-20\sqrt6+24+25+20\sqrt6+24+(25-24)`

Cancelling 20√6,

`\sf\longmapsto 25+24+25+24+(25-24)`

`\sf\longmapsto 98+1`

`\longmapsto\bf 99`

`\textsf{Therefore, the value of x²+y²+xy is 99.}\\`