Question

Urgent mate plz help ​

Answer 1

Question :-

If x = (√3 + √2)/(√3 - √2) and y = (√3 - √2)/(√3 + √2), find the value of x² + y²

Answer :-

Value of x² + y² = 98.

Solution :-

x = (√3 + √2)/(√3 - √2)

y = (√3 - √2)/(√3 + √2)

Finding the value of x + y

$\sf x + y = \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } + \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ \\ \sf = \dfrac{( \sqrt{3} + \sqrt{2} )( \sqrt{3} + \sqrt{2}) + ( \sqrt{3} - \sqrt{2})( \sqrt{3} - \sqrt{2})}{( \sqrt{3} - \sqrt{2})( \sqrt{3} + \sqrt{2}) } \qquad \left[taking\ lcm \right] \\ \\ \\ \sf = \dfrac{ {( \sqrt{3} + \sqrt{2} )}^{2} + {( \sqrt{3} - \sqrt{2} )}^{2} }{ {( \sqrt{3}) }^{2} - {( \sqrt{2}) }^{2} } \qquad \left[ \because (a + b)(a - b) = {a}^{2} - {b}^{2} \right] \\ \\ \\ \sf = \dfrac{ {( \sqrt{3})}^{2} + 2( \sqrt{3})( \sqrt{2}) + {( \sqrt{2} )}^{2} + {( \sqrt{3}) }^{2} - 2( \sqrt{3})( \sqrt{2}) + {( \sqrt{2}) }^{2} }{3 - 2} \qquad \left[ {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2} \ and \ {(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2} \right] \\ \\ \\ \sf = \dfrac{3 + 2 + 3 + 2}{1} \\ \\ \\ \rm x + y = 10$

Finding the value of xy

$\sf xy = \bigg(\dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \bigg) \bigg( \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \bigg) \\ \\ \\ \sf = \dfrac{( \sqrt{3} + \sqrt{2})( \sqrt{3} - \sqrt{2}) }{( \sqrt{3} - \sqrt{2})( \sqrt{3} + \sqrt{2} )} \\ \\ \\ \sf = \dfrac{ {( \sqrt{3})}^{2} - {( \sqrt{2}) }^{2} }{ {( \sqrt{3})}^{2} - {( \sqrt{2} )}^{2} } \qquad \left[ \because (a + b)(a - b) = {a}^{2} - {b}^{2} \right] \\ \\ \\ \sf = \dfrac{3 - 2}{3 - 2} \\ \\ \\ \rm xy = 1$

We know that

(x + y)² = x² + y² + 2xy

Here

• x + y = 10

• xy = 1

By substituting the values in the above identity

⇒ (10)² = x² + y² + 2(1)

⇒ 100 = x² + y² + 2

⇒ 100 - 2 = x² + y²

⇒ 98 = x² + y²

⇒ x² + y² = 98

Therefore the value of x² + y² = 98.

Answer 2

Answer:

plzzz follow me ***********************