Question

If $ x = 3 - 2\sqrt 2 $ , find the value of $ \sqrt x + \dfrac{1}{{\sqrt x }} $

Answer 1

Hint: To solve the given question, first we will write the given value of $ x $ and then try to solve with the $ x $ to find any of the parts of the given unfind expression. We will also rationalise the terms with square roots. We will solve until we will achieve the roots of $ x $ .

Complete step by step solution:
Given that:
 $ x = 3 - 2\sqrt 2 $
 $
   \Rightarrow \dfrac{1}{x} = \dfrac{1}{{3 - 2\sqrt 2 }} \\
   \Rightarrow \dfrac{1}{x} = \dfrac{1}{{3 - 2\sqrt 2 }} \times \dfrac{{3 + 2\sqrt 2 }}{{3 + 2\sqrt 2 }} \;
  $
In the above equation, to solve for the square roots we rationalised the terms with square roots.
 $
   \Rightarrow \dfrac{1}{x} = \dfrac{{3 + 2\sqrt 2 }}{{(3 - 2\sqrt 2 )(3 + 2\sqrt 2 }} \\
   \Rightarrow \dfrac{1}{x} = \dfrac{{3 + 2\sqrt 2 }}{{9 - 8}} = 3 + 2\sqrt 2 \;
  $
Now,
we have the values of $ x $ and $ \dfrac{1}{x} $ , so we can add both:
 $
  \therefore x + \dfrac{1}{x} = 3 - 2\sqrt 2 + 3 + 2\sqrt 2 \\
   \Rightarrow x + \dfrac{1}{x} = 6 \;
  $
we can add 2 in both L.H.S and R.H.S:
 $
   \Rightarrow x + \dfrac{1}{x} + 2 = 6 + 2 \\
   \Rightarrow {(\sqrt x )^2} + {(\dfrac{1}{{\sqrt x }})^2} + 2.\sqrt x .\dfrac{1}{{\sqrt x }} = 8 \\
   \Rightarrow {(\sqrt x + \dfrac{1}{{\sqrt x }})^2} = 8 \;
  $
Now, remove the square of L.H.S by doing the square root of R.H.S:
 $ \Rightarrow \sqrt x + \dfrac{1}{{\sqrt x }} = \pm \sqrt 8 $
 $ \therefore \sqrt x + \dfrac{1}{{\sqrt x }} = \pm 2\sqrt 2 $
Hence, the value of $ \sqrt x + \dfrac{1}{{\sqrt x }} $ is $ \pm 2\sqrt 2 $ .
So, the correct answer is “ $ \pm 2\sqrt 2 $ ”.

Note: We can get rid of a square root by squaring. (Or cube roots by cubing, etc) But Warning: this can sometimes create "solutions" which don't actually work when we put them into the original equation. So we need to Check!