Question

If x=√5+√3÷√5-√3 and y=√5-√3÷√5+√3,find the value of [x^2+y^2]

Answer 1

⇒ (√5 + √3 ÷ √5 - √3)²  ×  (√5 - √3 ÷ √5 + √3)

⇒(5 + 3 ÷ 5- 3)  × (5 - 3 ÷ 5 + 3)

⇒(8 ÷ 2)  (2 ÷ 8)

⇒4 × 0.25

⇒1

Answer 2

$The \ value \ of \ x^{2}+y^{2}=23$

Given:

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

$y=\frac{\sqrt{5} -\sqrt{3}}{\sqrt{5} + \sqrt{3}}$

Solution:

We know that,

$(x+y)^{2}=x^{2} + y^{2} + 2xy$

$x^{2} + y^{2}=(x+y)^{2} - 2xy$

Hence,

$x+y=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} + \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

$=\left(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right) + \left(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}} \times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}\right)$

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

$=\left(\frac{5+3}{5-2}\right) + \left(\frac{5-3}{5-2}\right)$

$=\frac{8}{2} + \frac{2}{2}$

$=\frac{8+2}{2}$

$\Rightarrow x + y = 5$

$(x+y)^{2}=25$

After that,

$2 x y=2\left(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)$

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

$=2\left(\frac{5-3}{5-2}\right)$

$=2\left(\frac{2}{2}\right)$

$\Rightarrow 2 x y=2$

Therefore,

$(x+y)^{2} - 2 x y = x^{2} + y^{2}$

$x^{2} + y^{2}=25-2$

$x^{2} + y^{2}=23$