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frozen [14]
2 years ago
11

If a = √3-√11 and b = 1 /a, then find a² - b²​

Mathematics
1 answer:
Serga [27]2 years ago
5 0

If b=\frac1a, then by rationalizing the denominator we can rewrite

b = \dfrac1{\sqrt3-\sqrt{11}} \times \dfrac{\sqrt3+\sqrt{11}}{\sqrt3+\sqrt{11}} = \dfrac{\sqrt3+\sqrt{11}}{\left(\sqrt3\right)^2-\left(\sqrt{11}\right)^2} = -\dfrac{\sqrt3+\sqrt{11}}8

Now,

a^2 - b^2 = (a-b) (a+b)

and

a - b = \sqrt3 - \sqrt{11} + \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{9\sqrt3 - 7\sqrt{11}}8

a + b = \sqrt3 - \sqrt{11} - \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{7\sqrt3 - 9\sqrt{11}}8

\implies a^2 - b^2 = \dfrac{\left(9\sqrt3 - 7\sqrt{11}\right) \left(7\sqrt3 - 9\sqrt{11}\right)}{64} = \boxed{\dfrac{441 - 65\sqrt{33}}{32}}

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