Question

If à=3-√5\3+√5 and b=3+√5/3-√5, fond a^2-b^2
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Answer 1

Solution :-

$a = \dfrac{3-\sqrt{5}}{3 + \sqrt{5}}$

$b = \dfrac{3+\sqrt{5}}{3 - \sqrt{5}}$

Now we will rationalise the denominators of a and b

For a

$a = \dfrac{3-\sqrt{5}}{3 + \sqrt{5}} \times \dfrac{3-\sqrt{5}}{3 - \sqrt{5}}$

$a= \dfrac{(3-\sqrt{5})^2}{3^2 - \sqrt{5}^2}$

$a = \dfrac{9-6\sqrt{5} + 5}{9 -5}$

$a = \dfrac{14 - 6\sqrt{5}}{4}$

$a = \dfrac{7 - 3\sqrt{5}}{2}$

For b

$b = \dfrac{3+\sqrt{5}}{3 - \sqrt{5}} \times \dfrac{3+\sqrt{5}}{3 + \sqrt{5}}$

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

$b = \dfrac{9+6\sqrt{5} + 5}{9 -5}$

$b = \dfrac{14 +6\sqrt{5}}{4}$

$b = \dfrac{7 + 3\sqrt{5}}{2}$

Now we have

$a = \dfrac{7 - 3\sqrt{5}}{2}$

and

$b = \dfrac{7 + 3\sqrt{5}}{2}$

Now

For a²

$a^2 =\left( \dfrac{7 - 3\sqrt{5}}{2}\right)^2$

$a^2 = \dfrac{49 - 42\sqrt{5} + 45}{4}$

$a^2 = \dfrac{94 - 42\sqrt{5}}{4}$

For b²

$b^2 =\left( \dfrac{7 + 3\sqrt{5}}{2}\right)^2$

$b^2 = \dfrac{49 + 42\sqrt{5} + 45}{4}$

$b^2 = \dfrac{94 + 42\sqrt{5}}{4}$

Now a² - b²

$= \dfrac{94 - 42\sqrt{5}}{4} - \dfrac{94 + 42\sqrt{5}}{4}$

$= \dfrac{(94 -94) - 42\sqrt{5}- 42\sqrt{5}}{4}$

$= \dfrac{ -84\sqrt{5}}{4}$

$= -21\sqrt{5}$