Question

If x = 3 + root2/3 - root2 and y = 3-root2/ 3+root2 , find the value of x2 - y2

Answer 1

Answer:

Step-by-step explanation:

X =(√3+√2)/(√3-√2)

By taking conjugate, we get

X  = (√3+√2) (√3+√2) /(√3-√2) (√3+√2)

    =(√3+√2)²/(√3²-√2²)

    =3+2√6+2 / (3-2) = 5+2√6

Y   = (√3-√2)/(√3+√2)

By taking conjugate we get

Y    = (√3-√2)²/ (√3+√2)(√3-√2)

      = 3-2√6+2/√3²-√2²

       = 5-2√6/(3-2) = 5-2√6

X²     = (5+2√6)² = 25+20√6+24 = 49+20√6

Y²      = (5-2√6)² = 25-20√6+24  = 49-20√6

X²+Y² =  49+20√6+49-20√6

X²+Y² = 98

Hope it helps you...

Answer 2

$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ \\ x + y = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } + \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ x + y = \frac{( { \sqrt{3} + \sqrt{2} })^{2} + ( { \sqrt{3} - \sqrt{2} })^{2} }{3 - 2} \\ \\ x + y = 2(3 + 2) = 10 \\ \\ \\ x - y = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } - \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ x - y = \frac{( { \sqrt{3} + \sqrt{2} })^{2} - ( { \sqrt{3} - \sqrt{2} })^{2} }{3 - 2} \\ \\ x - y = 4 \sqrt{3} \sqrt{2} = 4 \sqrt{6} \\ \\ \\ {x}^{2} - {y}^{2} = (x + y)(x - y) \\ = 10 \times 4 \sqrt{6} \\ = 40 \sqrt{6}$