Finding the value of $\sqrt[4]{(4+\sqrt7)^{-1}}\sqrt{1+\sqrt7}$ with other approaches

Question

It is a problem from a timed exam,

What is the value of $\sqrt[4]{(4+\sqrt7)^{-1}}\sqrt{1+\sqrt7}$ ?

$1)1\qquad\qquad2)\sqrt[4]2\qquad\qquad3)2\qquad\qquad4)2\sqrt[4]2$

I solved it with two approaches.

First approach,

$$\sqrt[4]{\frac{1}{4+\sqrt7}}\times\sqrt{1+\sqrt7}=\sqrt[4]{\frac{2}{(1+\sqrt7)^2}}\times\sqrt{1+\sqrt7}=\sqrt[4]2$$

Second approach, $$\sqrt[4]{\frac{1}{4+\sqrt7}}\times\sqrt{1+\sqrt7}=\sqrt[4]{\frac{1}{4+\sqrt7}}\times\sqrt[4]{8+2\sqrt7}=\sqrt[4]{\frac{2(4+\sqrt7)}{4+\sqrt7}}=\sqrt[4]2$$ I'm wondering is it possible to solve this problem with other efficient approaches?

Answer 1

Personally I would have gone for $x^4=\dfrac{(1+\sqrt{7})^2}{4+\sqrt{7}}=\dfrac{8+2\sqrt{7}}{4+\sqrt{7}}=2$

In which you discover along the way that $(1+\sqrt{7})^2$ comes out perfect for a simplification, while I feel that in your approach it is prerequisite.

Also I do not have to carry on drawing all these outer roots symbols...

Answer 2

Your methods are perfectly alright even for a timed examination.

In particular, I liked the second solution. Another possible way to achieve the same would be rationalising the denominator inside the first radical.

Answer 3

If we call $ \ u \ = \ 4 + \sqrt7 \ \ , $ then we have $ \ 1 + \sqrt7 \ = \ u - 3 \ $ and $ \ \frac{1}{u} \ = \ \frac{4 - \sqrt7}{9} \ \ . $ Consequently,

$$\sqrt[4]{(4+\sqrt7)^{-1}} \ · \ \sqrt{1+\sqrt7} \ \ = \ \ \sqrt[4]{\frac{(u \ - \ 3)^2}{u}} \ \ = \ \ \sqrt[4]{ \ u \ - \ 6 \ + \ \frac{9}{u} \ }$$ $$ = \ \ \sqrt[4]{ \ (4 + \sqrt7) \ - \ 6 \ + \ 9· \left(\frac{4 - \sqrt7}{9} \right) \ } \ \ = \ \ \sqrt[4]{ \ 4 \ - \ 6 \ + \ 4 \ } \ \ = \ \ \sqrt[4]{ 2 } \ \ . $$