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Question

If $$x=\displaystyle\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$$ and $$y=\displaystyle\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$$ , find the value of $${x}^{2}+{y}^{2}-6xy$$.


Solution

Given that

$$x= \dfrac{\sqrt5 - \sqrt3}{\sqrt5 +\sqrt3}$$

$$= \dfrac{(\sqrt5 - \sqrt3)(\sqrt5-\sqrt3)}{(\sqrt5 +\sqrt3)(\sqrt5-\sqrt3)}$$

$$= \dfrac{8-2\sqrt{15} }{2}$$

$$x= 4-\sqrt{15}$$

$$x^2= 31-8\sqrt{15}$$  .....................................(1)


$$y= \dfrac{\sqrt5 + \sqrt3}{\sqrt5 -\sqrt3}$$

$$= \dfrac{(\sqrt5 + \sqrt3)(\sqrt5+\sqrt3)}{(\sqrt5 +\sqrt3)(\sqrt5-\sqrt3)}$$

$$= \dfrac{8+2\sqrt{15} }{2}$$

$$y= 4+\sqrt{15}$$

$$y^2= 31+8\sqrt{15}$$  .....................................(2)


$$xy= [4 - \sqrt{15}] \times [4 +\sqrt{15}]$$

$$ \Rightarrow xy= 1$$  ........................................(3)

Now we have to find out the value of given expression

$$x^{2}+y^2 -6xy =  (31-8\sqrt{15})+(31+8\sqrt{15})-6$$

                          $$=62-6 =56$$

Hence, $$56$$ is the answer

Mathematics

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