If x= (2- 5)/(2+\sqrt5)/ (2+ \sqrt5) and y = (2+ 5)/(2−\sqrt5) / (2- \sqrt5) then what will be the value of x^ 2 - y^2

asked by guest
on Mar 21, 2025 at 12:21 am



MathBot Answer:

x2−y2=−(2+5)2(2−5)2+(2−5)2(2+5)2{x}^{2} - {y}^{2} = - \frac{\left(2 + \sqrt{5}\right)^{2}}{\left(2 - \sqrt{5}\right)^{2}} + \frac{\left(2 - \sqrt{5}\right)^{2}}{\left(2 + \sqrt{5}\right)^{2}}

x2−y2=(2−55+2)2−(5+22−5)2=(2−55+2)2−(−5−25−2)2=(2−5)2(5+2)2−(−5−25−2)2=(2−5)2(5+2)2−(−5−2)2(5−2)2=−(−5−2)2(5−2)2+9−4545+9=9−4545+9−−45−945−9=45+945−9+9−4545+9=−(2+5)2(2−5)2+(2−5)2(2+5)2\begin{aligned}x^{2} - y^{2}&=\left(\frac{2 - \sqrt{5}}{\sqrt{5} + 2}\right)^{2} - \left(\frac{\sqrt{5} + 2}{2 - \sqrt{5}}\right)^{2}\\&=\left(\frac{2 - \sqrt{5}}{\sqrt{5} + 2}\right)^{2} - \left(\frac{- \sqrt{5} - 2}{\sqrt{5} - 2}\right)^{2}\\&=\frac{\left(2 - \sqrt{5}\right)^{2}}{\left(\sqrt{5} + 2\right)^{2}} - \left(\frac{- \sqrt{5} - 2}{\sqrt{5} - 2}\right)^{2}\\&=\frac{\left(2 - \sqrt{5}\right)^{2}}{\left(\sqrt{5} + 2\right)^{2}} - \frac{\left(- \sqrt{5} - 2\right)^{2}}{\left(\sqrt{5} - 2\right)^{2}}\\&=-\frac{\left(- \sqrt{5} - 2\right)^{2}}{\left(\sqrt{5} - 2\right)^{2}} + \frac{9 - 4 \sqrt{5}}{4 \sqrt{5} + 9}\\&=\frac{9 - 4 \sqrt{5}}{4 \sqrt{5} + 9} - \frac{- 4 \sqrt{5} - 9}{4 \sqrt{5} - 9}\\&=\frac{4 \sqrt{5} + 9}{4 \sqrt{5} - 9} + \frac{9 - 4 \sqrt{5}}{4 \sqrt{5} + 9}\\&=- \frac{\left(2 + \sqrt{5}\right)^{2}}{\left(2 - \sqrt{5}\right)^{2}} + \frac{\left(2 - \sqrt{5}\right)^{2}}{\left(2 + \sqrt{5}\right)^{2}}\end{aligned}


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