Question

How do you simplify $ \sqrt{3-2\sqrt{2}} $ ?

Answer 1

Hint: We first try to form the equation $ 3-2\sqrt{2} $ into a square form. We apply the identity of squares where $ {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} $ . We take the irrational as the $ -2ab $ part. After getting the square form we take the root value and two solutions.

Complete step by step answer:
We have been given to find the square root value of $ 3-2\sqrt{2} $ .
Therefore, we try to form the algebraic equation $ 3-2\sqrt{2} $ into a full square form.
There is an irrational number $ \sqrt{2} $ in the equation.
The square form we take will be a square of subtraction of two numbers $ a $ and $ b $ .
The formula being as $ {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} $ .
We will try to form the $ -2ab $ part for the irrational part.
We can form the equation as $ 3-2\sqrt{2}=2+1-2\sqrt{2} $ .
We take the terms 2 and 1 as the squares.
2 will be considered as the square of $ \sqrt{2} $ . For the second part 1 will be considered as the square of 1.
Therefore, $ 2+1-2\sqrt{2}={{\left( \sqrt{2} \right)}^{2}}-2\times 1\times \sqrt{2}+{{1}^{2}} $ .
Equating with the formula $ {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} $ , we get $ {{\left( \sqrt{2} \right)}^{2}}-2\times 1\times \sqrt{2}+{{1}^{2}}={{\left( \sqrt{2}-1 \right)}^{2}} $ .
Now we need to find the square root of the $ 3-2\sqrt{2} $ which will be equal to $ \sqrt{{{\left( \sqrt{2}-1 \right)}^{2}}} $ .
Now square root value will give two answers where the values are the same but their signs being opposite to each other.
Therefore, $ \sqrt{3-2\sqrt{2}}=\sqrt{{{\left( \sqrt{2}-1 \right)}^{2}}}=\pm \left( \sqrt{2}-1 \right) $ .
The simplified form of $ \sqrt{3-2\sqrt{2}} $ is $ \pm \left( \sqrt{2}-1 \right) $ .

Note:
 we can also argue about the answer being only positive as the sign of the root is already provided where it is given to find the value of $ \sqrt{3-2\sqrt{2}} $. But it is always suggested to get two values for the root value instead of one.