Question

plz solve this question​

Answer 1

QUESTION:

$\sf If \ \dfrac{2+\sqrt{5}}{2-\sqrt{5}} = x\ and\ \dfrac{2-\sqrt{5}}{2+\sqrt{5}} = y\ find$

$\sf the\ value\ of\ x^{2} - y^{2}$

ANSWER:

$\sf The\ value\ of\ x^{2} - y^{2} = 144 \sqrt{5}$

GIVEN:

$\sf \dfrac{2+\sqrt{5}}{2-\sqrt{5}} = x\ and\ \dfrac{2-\sqrt{5}}{2+\sqrt{5}} = y$

TO FIND:

$\sf The\ value\ of\ x^{2} - y^{2}$

EXPLANATION:

$\boxed{\bold{\large{\gray{x^2 - y^2= (x+y)(x-y)}}}}$

$\sf x + y= \dfrac{2+\sqrt{5}}{2-\sqrt{5}} + \dfrac{2-\sqrt{5}}{2+\sqrt{5}}$

$\sf x + y = \dfrac{(2+\sqrt{5})^{2} + (2-\sqrt{5})^{2} }{2^{2} -(\sqrt{5})^{2} }$

(A + B)² = A² + 2AB + B²

(A - B)² = A² - 2AB + B²

(A + B)² + (A - B)² = A² + 2AB + B² + A² - 2AB + B²

(A + B)² + (A - B)² = 2A² + 2B²

$\sf x + y= \dfrac{2(2^{2} +(\sqrt{5})^{2} ) }{2^{2} -(\sqrt{5})^{2} }$

$\sf x + y= \dfrac{2(4 +5) }{4 - 5 }$

$\sf x + y= \dfrac{2(9) }{ - 1 }$

$\sf x + y= - 18$

$\sf x - y= \dfrac{2+\sqrt{5}}{2-\sqrt{5}} - \dfrac{2-\sqrt{5}}{2+\sqrt{5}}$

$\sf x - y = \dfrac{(2+\sqrt{5})^{2} - (2-\sqrt{5})^{2} }{2^{2} -(\sqrt{5})^{2} }$

(A + B)² = A² + 2AB + B²

(A - B)² = A² - 2AB + B²

(A + B)² - (A - B)² = A² + 2AB + B² - A² + 2AB - B²

(A + B)² - (A - B)² = 4 AB

$\sf x - y = \dfrac{4(2 \sqrt{5})}{2^{2} -(\sqrt{5})^{2} }$

$\sf x - y = \dfrac{8\sqrt{5}}{4 - 5}$

$\sf x - y = - 8\sqrt{5}$

$\sf x^{2} - y^{2} = - 18( - 8 \sqrt{5} )$

$\sf x^{2} - y^{2} =144 \sqrt{5}$

$\sf Hence \ the\ value \ of \ x^{2} - y^{2} =144 \sqrt{5}$