Question

If x=√3+√2÷√3-√2 and y=√3-√2÷√3+√2 then find the value of x×x+y×y

Answer 1

rationalising the denominator

x = ( √3 + √2 ÷ √3 - √2 )× (√3 + √2 / √3 + √2)

= 3+ 2√6 +2 / 3 - 2

= 5+2√6

y= (√3 - √2 ÷ √3 + √2) × (√3 - √2 / √3 - √2)

= 3 - 2√6 + 2 / 3 - 2

= 5 - 2√6

now

x^2 + y^2

(5+2√6)^2 + (5-2√6)^2

25 + 20√6 + 24 + 25 - 20√6 + 24

49 + 20√6. +. 49 - 20√6

98

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

$\tt{\rightarrow x=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}$

Rationalize it

$\tt{\rightarrow x=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

$\tt{\rightarrow x=\dfrac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}}$

$\tt{\rightarrow x=\dfrac{3+2+2\sqrt{3}\sqrt{2}}{3-2}}$

x = 5 + 2√6

When we solve y in similar way we get

y = 5 - √6

Therefore,

x² + y²

= (5 + 2√6)² + (5 - 2√6)²

= {25 + 2 × 5 × 2√6 + (2√6)²} + {25 - 2 × 5 × 2√6 + (2√6)²)

= (25 + 20√6 + 24) + (25 - 20√6 + 24)

= 49 + 20√6 + 49 - 20√6

= 98