Question

if a=1/3-√11 and b=1/a, then find a^2 - b^2

Answer 1

a =√11/3b = 1/a = 1/(√11/3) = 3/√11a2 - b2 = (√11/3)2 -(3/√11)2 ={(√11)2/(3)2} - {(3)2/(√11)2}= (11/9) - (9/11) = (121-81)/99= 40/99

Answer 2

The answer is

        $a^{2} + b^{2} = \frac{15\sqrt{11}- 30 }{2}$

Given:  

$a = \frac{1}{3 - \sqrt{11} }$   and $b = \frac{1}{a}$  

To find:

The value of a² - b²

Solution:        

Given  $a = \frac{1}{3 - \sqrt{11} }$  

To rationalize the denominator multiply and divide with (3+√11)

⇒ $a = \frac{1}{3 - \sqrt{11} } (\frac{3+\sqrt{11} }{3+ \sqrt{11} } )$

⇒  $\frac{3+\sqrt{11} }{3^{2} - (\sqrt{11} )^{2} } = \frac{3+\sqrt{11} }{9 -11 }$

⇒  $a = \frac{3+\sqrt{11} }{-2}$  

$a^{2} = \frac{3+\sqrt{11} }{-2}(\frac{3+\sqrt{11} }{-2})$  $= \frac{9+11 + 6\sqrt{11} }{4}$

   $= \frac{20 + 6\sqrt{11} }{4}$  $= \frac{10 + 3\sqrt{11} }{2}$  

⇒ $a^{2} = \frac{10 + 3\sqrt{11} }{2}$  ---- (1)

Given  $b = \frac{1}{a}$  

⇒ $b = \frac{1}{\frac{1}{3 - \sqrt{11} } } = 3 - \sqrt{11}$  

⇒ $b^{2} = (3 - \sqrt{11})^{2}$ = $= (3 - \sqrt{11})^{2} = 9 + 11 - 6\sqrt{11}$

⇒ $b^{2} = 20 - 6\sqrt{11}$-----(2)

From (1) and (2)

$a^{2} + b^{2} = \frac{10 + 3\sqrt{11} }{2} - ( 20 - 6\sqrt{11})$  

$= \frac{10 + 3\sqrt{11}- 40 +12\sqrt{11} }{2}$  

$= \frac{15\sqrt{11}- 30 }{2}$  

⇒ $a^{2} + b^{2} = \frac{15\sqrt{11}- 30 }{2}$

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