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Mathematics

If 2+52−5=x and 2−52+5=y\dfrac{2 + \sqrt{5}}{2 - \sqrt{5}} = x \text{ and } \dfrac{2 - \sqrt{5}}{2 + \sqrt{5}} = y; find the value of x2 - y2.

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Answer

Given,

x = 2+52−5\dfrac{2 + \sqrt{5}}{2 - \sqrt{5}}

Rationalizing,

⇒x=2+52−5×2+52+5=(2+5)222−(5)2=22+(5)2+2×2×54−5=4+5+45−1=−(9+45).⇒x2=[−(9+45)]2=92+(45)2+2×9×45=81+80+725=161+725.\Rightarrow x = \dfrac{2 + \sqrt{5}}{2 - \sqrt{5}} \times \dfrac{2 + \sqrt{5}}{2 + \sqrt{5}} \\[1em] = \dfrac{(2 + \sqrt{5})^2}{2^2 - (\sqrt{5})^2} \\[1em] = \dfrac{2^2 + (\sqrt{5})^2 + 2\times 2 \times \sqrt{5}}{4 - 5} \\[1em] = \dfrac{4 + 5 + 4\sqrt{5}}{-1} \\[1em] = -(9 + 4\sqrt{5}). \\[1em] \Rightarrow x^2 = [-(9 + 4\sqrt{5})]^2 \\[1em] = 9^2 + (4\sqrt{5})^2 + 2\times 9 \times 4\sqrt{5} \\[1em] = 81 + 80 + 72\sqrt{5} \\[1em] = 161 + 72\sqrt{5}.

Given,

y = 2−52+5\dfrac{2 - \sqrt{5}}{2 + \sqrt{5}}

Rationalizing,

⇒y=2−52+5×2−52−5=(2−5)222−(5)2=22+(5)2−2×2×54−5=4+5−45−1=−(9−45).⇒y2=[−(9−45)]2=92+(45)2−2×9×45=81+80−725=161−725.\Rightarrow y = \dfrac{2 - \sqrt{5}}{2 + \sqrt{5}} \times \dfrac{2 - \sqrt{5}}{2 - \sqrt{5}} \\[1em] = \dfrac{(2 - \sqrt{5})^2}{2^2 - (\sqrt{5})^2} \\[1em] = \dfrac{2^2 + (\sqrt{5})^2 - 2\times 2 \times \sqrt{5}}{4 - 5} \\[1em] = \dfrac{4 + 5 - 4\sqrt{5}}{-1} \\[1em] = -(9 - 4\sqrt{5}). \\[1em] \Rightarrow y^2 = [-(9 - 4\sqrt{5})]^2 \\[1em] = 9^2 + (4\sqrt{5})^2 - 2\times 9 \times 4\sqrt{5} \\[1em] = 81 + 80 - 72\sqrt{5} \\[1em] = 161 - 72\sqrt{5}.

Substituting values of x2 and y2, we get :

⇒x2−y2=(161+725)−(161−725)=161−161+725+725=1445.\Rightarrow x^2 - y^2 = (161 + 72\sqrt{5}) - (161 - 72\sqrt{5}) \\[1em] = 161 - 161 + 72\sqrt{5} + 72\sqrt{5} \\[1em] = 144\sqrt{5}.

Hence, x2 - y2 = 1445144\sqrt{5}.

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