Question

if x=2-√3/2+√3 and y=2+√3/2-√3 find x²-y²​

Answer 1

Answer:

Answer is in Attachment....................

Answer 2

Given:-

• x = $\sf{\dfrac{2 - \sqrt{3}}{2 + \sqrt{3}}}$
• y = $\sf{\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}}$

To Find:-

• The value of x² - y²

Solution:-

We have,

x = $\sf{\dfrac{2 - \sqrt{3}}{2 + \sqrt{3}}}$

By Rationalizing the denominator

$\sf{\dfrac{2 - \sqrt{3}}{2 + \sqrt{3}}\times \dfrac{2 - \sqrt{3}}{2 - \sqrt{3}}}$

= $\sf{\dfrac{(2 - \sqrt{3})^2}{(2)^2 - (\sqrt{3})^2}}$

By applying the identity,

(a - b)² = a² - 2ab + b²

= $\sf{\dfrac{(2)^2 - 2\times 2\times\sqrt{3} + (\sqrt{3})^2}{4 - 3}}$

= $\sf{\dfrac{4 - 4\sqrt{3} + 3}{1}}$

= $\sf{7 - 4\sqrt{3}}$

Therefore x = (7 - 4√3)

Now,

We also have,

y = $\sf{\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}}$

By Rationalizing the denominator,

$\sf{\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}\times \dfrac{2 + \sqrt{3}}{2 + \sqrt{3}}}$

= $\sf{\dfrac{(2 + \sqrt{3})^2}{(2)^2 - (\sqrt{3})^2}}$

By applying the identity,

(a + b)² = a² + 2ab + b²

= $\sf{\dfrac{(2)^2 + 2\times 2 \times \sqrt{3} + (\sqrt{3})^2}{4 - 3}}$

= $\sf{\dfrac{4 + 4\sqrt{3} + 3}{1}}$

= $\sf{7 + 4\sqrt{3}}$

Therefore y = 7 + 4√3

Now,

We were asked to find the value of x² - y²

Hence putting the values of x and y,

$\sf{x^2 - y^2 = (7 - 4\sqrt{3})^2 - (7 + 4\sqrt{3})^2}$

By applying the property

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

$\sf{(7 - 4\sqrt{3})^2 - (7 + 4\sqrt{3})^2}$

$\sf{=[(7)^2 - 2\times 7 \times 4\sqrt{3} + (4\sqrt{3})^2] - [(7)^2 + 2\times 7 \times 4\sqrt{3} + (4\sqrt{3})^2}$

= $\sf{[49 - 56\sqrt{3} + 48] - [49 + 56\sqrt{3} + 48]}$

= $\sf{49 - 56\sqrt{3} + 48 - 49 - 56\sqrt{3} - 48}$

= $\sf{\cancel{49} - 56\sqrt{3} + \cancel{48} - \cancel{49} - 56\sqrt{3} - \cancel{48}}$

= $\sf{-56\sqrt{3} - 56\sqrt{3}}$

= $\sf{-112\sqrt{3}}$

Therefore,

The value of x² - y² is -112√3.

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