Question

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

Answer 1

Answer:

The answer is $\dfrac{15\sqrt{11}-30}{2}$

Step-by-step explanation:

Given

        $a=\dfrac{1}{3-\sqrt{11}}$

Rationalize the denominator

      $\Rightarrow a=\dfrac{1}{3-\sqrt{11}}\times \dfrac{3+\sqrt 11}{3+\sqrt{11}}$

              $=\dfrac{3+\sqrt{11}}{9-11}=\dfrac{3+\sqrt{11}}{-2}$

    $\Rightarrow a^2=\dfrac{9+11+6\sqrt{11}}{4}=\dfrac{20+6\sqrt{11}}{4}=\dfrac{10+3\sqrt{11}}{2}$...(1)

Also given

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

       $\Rightarrow b^2=(3-\sqrt{11})^2=9+11-6\sqrt{11}=20-6\sqrt{11}$

Therefore

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

                   $=\dfrac{10+3\sqrt{11}-40+12\sqrt{11}}{2}=\dfrac{15\sqrt{11}-30}{2}$

PS: I suspect some error in this questions