Question

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

Answer 1

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

`b = \frac1{\sqrt3-√(11)} * (\sqrt3+√(11))/(\sqrt3+√(11)) = (\sqrt3+√(11))/(\left(\sqrt3\right)^2-\left(√(11)\right)^2) = -\frac{\sqrt3+√(11)}8`

Now,

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

and

`a - b = \sqrt3 - √(11) + \frac{\sqrt3 + √(11)}8 = \frac{9\sqrt3 - 7√(11)}8`

`a + b = \sqrt3 - √(11) - \frac{\sqrt3 + √(11)}8 = \frac{7\sqrt3 - 9√(11)}8`

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