Question

if x is equal to root 5 + 1 upon root 5 minus 1 and Y is equal to root 5 minus 1 upon root 5 + 1 then find the value of x square + Y square

Answer 1

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Answer 2

Answer:

x² + y² = 7

Step-by-step explanation:

Give values of x and y are

$x=\frac{\sqrt{5}+1}{\sqrt{5}-1}\:\:and\:\:y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$

To find: x² + y²

first we find value of x and y by rationalizing the denominator

Consider,

$x=\frac{\sqrt{5}+1}{\sqrt{5}-1}$

$x=\frac{\sqrt{5}+1}{\sqrt{5}-1}\times\frac{\sqrt{5}+1}{\sqrt{5}+1}$

$x=\frac{(\sqrt{5}+1)^2}{(\sqrt{5}-1)(\sqrt{5}+1)}$

$x=\frac{(\sqrt{5})^2+1+2\times\sqrt{5}}{(\sqrt{5})^2-1}$

$x=\frac{5+1+2\times\sqrt{5}}{5-1}$

$x=\frac{6+2\times\sqrt{5}}{4}$

$x=\frac{3+\sqrt{5}}{2}$

$y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$

$x=\frac{\sqrt{5}-1}{\sqrt{5}+1}\times\frac{\sqrt{5}-1}{\sqrt{5}-1}$

$x=\frac{(\sqrt{5}-1)^2}{(\sqrt{5}+1)(\sqrt{5}-1)}$

$x=\frac{(\sqrt{5})^2+1-2\times\sqrt{5}}{(\sqrt{5})^2-1}$

$x=\frac{5+1-2\times\sqrt{5}}{5-1}$

$x=\frac{6-2\times\sqrt{5}}{4}$

$x=\frac{3-\sqrt{5}}{2}$

Finally Consider,

$x^2+y^2$

$=(\frac{3+\sqrt{5}}{2})^2+(\frac{3-\sqrt{5}}{2})^2$

$=\frac{(3+\sqrt{5})^2}{4}+\frac{(3-\sqrt{5})^2}{4}$

$=\frac{9+(\sqrt{5})^2+2\times3\times\sqrt{5}+9+(\sqrt{5})^2-2\times3\times\sqrt{5}}{4}$

$=\frac{9+5+9+5}{4}$

$=\frac{28}{4}$

$=7$

Therefore, x² + y² = 7