Question

if a = 3 +2 root 2 , then find the value of root2 - 1÷roota
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Answer 1

Answer:

The answer is $3-2\sqrt 2$

Step-by-step explanation:

Given $a=3+2\sqrt 2$

             $=2+1+2\sqrt 2\\=(\sqrt 2)^2+1^2+2(\sqrt 2)(1)\\=(\sqrt 2+1)^2$

 $\Rightarrow \sqrt a=\sqrt 2+1$

Therefore

     $\dfrac{\sqrt 2-1}{\sqrt a}=\dfrac{\sqrt 2-1}{\sqrt 2+1}$

                 $=\dfrac{\sqrt 2-1}{\sqrt 2+1}\times \dfrac{\sqrt 2-1}{\sqrt2 -1}\\\\=\dfrac{(\sqrt2)^2+1^2-2\sqrt 2}{(\sqrt 2)^2-1^2}\\\\=\dfrac{2+1-2\sqrt 2}{2-1}=3-2\sqrt 2$