Tricky Radicals

Let $x=\sqrt { 3+2\sqrt { 2 } } -\sqrt { 3-2\sqrt { 2 } }$ . Then we have

${ x }^{ 2 }={ \left( \sqrt { 3+2\sqrt { 2 } } -\sqrt { 3-2\sqrt { 2 } } \right) }^{ 2 }\\ =3+2\sqrt { 2 } +3-2\sqrt { 2 } -2\sqrt { 3+2\sqrt { 2 } } \sqrt { 3-2\sqrt { 2 } } \\ =6-2\sqrt { \left( 3+2\sqrt { 2 } \right) \left( 3-2\sqrt { 2 } \right) }$

Notice the difference of two squares within the root: ${ x }^{ 2 }=6-2\sqrt { 9-8 } =4$

So we have $x=\pm 2$ . We can disregard $-2$ because $\sqrt { 3+2\sqrt { 2 } } >\sqrt { 3-2\sqrt { 2 } }$ so it is clear that our answer will be positive.

$x=\large \color{#20A900}{\boxed{2}}$

$3+2\sqrt{2}=(\sqrt{2}+1)^{2}$ and $3-2\sqrt{2}=(\sqrt{2}-1)^{2}$ Now on to the problem: $\sqrt {(\sqrt{2}+1)^{2}}-\sqrt {(\sqrt{2}-1)^2}\\ =\sqrt {2}+1-(\sqrt{2}-1)\\=\boxed {2}$

We power the given expression by 2 to get: $(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}})^{2}=4$ So we get the result: $R=\boxed{2}$

This problem would be much easier if we understand this pattern:

$(\sqrt{a}+\sqrt{b})^2 = a + b + 2\sqrt{ab}$

Therefore, we have:

$\sqrt{a + b + 2\sqrt{ab}} = \sqrt{a}+\sqrt{b}$

Now, can you see that:

$\sqrt{3+2\sqrt{2}} = \sqrt{2} + \sqrt{1}$

and

$\sqrt{3-2\sqrt{2}} = \sqrt{2} - \sqrt{1}$ *

?

Once we see that, it's easy to calculate:

$\sqrt{3+2\sqrt{2}} - \sqrt{3-2\sqrt{2}} = (\sqrt{2} + \sqrt{1}) - (\sqrt{2} - \sqrt{1})$

$= 2\sqrt{1}$ $= \fbox{2}$

*) The $\sqrt{2}$ must be in the front because the result of the square root must be possitive

If $x_{1}$ , $x_{2}$ are 2 real roots of the quadratic equation $x^{2}-ax+\frac{b^{2}c}{4} \left(a,b,c>0\right)$ then we have $a\pm b\sqrt{c}=\left(\sqrt{x_{1}}\pm \sqrt{x_{2}}\right)^{2}$

Try this on your calculator: √(3+2×√2)−√(3−2×√2)

Let a=3, b=2√2 then we are looking for √(a+b)-√(a-b)=x

a+b-2√(a^2-b^2 )+a-b=x^2

2a-2√(a^2-b^2 )=x^2

2(a-√(a^2-b^2))=x^2

Now substitute a and b to get:

2(3-√(9-8))=x^2

4=x^2

x=2 (since x must be positive)

2 times square root of 2 = 2 So square root of 5 minus square root of 1 Equals square root of 4 Equals 2

Let a = √(3+2√2) and b = √(3-2√2) We observe that ab=1; Also a + b=6 (√a+ √b)( √a- √b) = a – b ………(1) a – b = √(〖(a+b)〗^2-4ab)=4√2 ……(2) (√a+ √b)2 = a + b + 2√ab
=8 …….(3)

From 1 , 2, 3 √a- √b = 2