Introductory Olympiad Algebra - Nested Square Roots

We work from the inner square root. We guess that $6 - 4 \sqrt{2}$ is a perfect square of the form $a + b \sqrt{c}$ . Since

$6 - 4 \sqrt{2} = ( a + b \sqrt{c} ) ^2 = a^2 + b^2 c - 2 ab \sqrt{c} ,$

we see that $6 = a^2 + b^2 c, -4 = -2ab , 2=c$ . This gives us $a = 2, b = - 1, c = 2$ . Hence $6 - 4 \sqrt{2} = (2 - \sqrt{2})^2$ . Since $2 - \sqrt{2} > 0$ , thus $\sqrt{ 6 - 4 \sqrt{2}} = (2 - \sqrt{2} )$ .

In a similar manner, we guess that the outer square root $23 - 6( 2 - \sqrt{2} ) = 11 + 6 \sqrt{2}$ also has the form $x + y \sqrt{z}$ . Since

$11 + 6 \sqrt{2} = (x + y \sqrt{z} )^2 = x^2 + y^2z + 2xy \sqrt{z} ,$

we see that $11 = x^2 + y^2 z , 6 = 2xy, 2 = z$ . Then, $23 - 6 \sqrt{ 6 - 4 \sqrt{2}} = 23 - 6 ( 2 - \sqrt{2}) = 11 + 6 \sqrt{2} = (3 + \sqrt{2})^2$ Hence, the expression is equal to $3 + \sqrt{2}$ . This gives us $x = 3, y = 1, z = 2$ , hence $11 + 6 \sqrt{2} = ( 3 + \sqrt{2} ) ^2$ . Since $3 + \sqrt{2} > 0$ , hence the answer is $3 + \sqrt{2}$ .

Working from the inner root... Sq.root(6-2.root2)=sq.root[(2)^2+(root2)^2-2.2.root2]=sq.root(2-root2)^2=(2-root2)..next we have sq.root(11+3root2) which in similer way turns out to be 3+root(2)

Olympiad problems are much more difficult

Innermost sqr is smaller than one

Entire inner expression is 20ish

Solution is (most probably) 4 something

first make 6-4X2^1/2 in the form of a^2 +b^2-2ab which makes it (2-2^1/2)^2. Under the square root, it becomes 2-2^1/2. Using the same principle, the rest of the sum can be solved.

6-4((2)^(1/2)) =(2-(2)^(1/2))^2 and then the relation becomes=(11+6 2^(1/2))= 9+2+(6 2(1/2)) so solution is 3+2^(1/2)