Explain how sqrt(4+2sqrt(3)) becomes sqrt(3)+1?

Question

I am doing some math problems at home and the book im working from is terrible at explaining, it just says the question and answer, but I have no idea how sqrt(4+2sqrt(3)) simplifies to sqrt(3)+1? I have a few ideas but nothing I am too confident in thinking is right

Could someone explain in detail, I tried looking up on nested radicals but almost all focus highly on infinite sums and Wikipedia was confusing. Lauder's Algorithm is one hell of a pain in the *** to find too.

Any help is greatly appreciated thanks

PS. I asked this 2 hours ago but YA ate the question :/ or I just cant see it myself, sorry if its a duplicate

Answer 1

Really weird problem. I was amazed that it was not a typo. I didn't think this could possibly be correct until I checked it with a calculator.



Here's what's going on with it. You need to reexpress 4 + 2√3 as a perfect square.



4 + 2sqrt(3) = 3 + 2sqrt(3) + 1 = (sqrt(3) + 1)^2

So sqrt(4 + 2sqrt(3)) = sqrt( (sqrt(3) + 1)^2 ) = sqrt(3) + 1



You could also square √3 + 1 to see what it gives you:

(√3 + 1)^2 = 3 + 2√3 + 1 = 4 + 2√3

Answer 2

√(4 + 2√3) = √3 +1

4 + 2√3 = (√3 + 1)^2

4 + 2√3 = 3 + 2√3 + 1

4 + 2√3 = 4 + 2√ 3

Answer 3

This question can be derived from (sqrt(a)+sqrt(b))^2=a+b+2*sqrt(ab)

Squaring both sides, then we got same form with your question: sqrt(a)+sqrt(b)=sqrt(a+b+2*sqrt(ab))

From the original question, we got a+b=4 and ab=3

By using substitution, a=4-b and (4-b).b=3

Evaluating the last equation yields 4b-b^2-3=0 or b^2-4b+3=0, then we got b=3 or b=1.

So a=1 or a=3. Substituting the value of a and b to the original question, so we get:

sqrt(4+2sqrt(3))=sqrt(1)+sqrt(3) = 1+sqrt(3)

That's the answer :)

Answer 4

Jane: If sqrt(4 + 2sqrt3) = sqrt3 + 1 then if I square both sides they must still be the same. Squaring the LHS just gets rid of the outside sqrt, leaving 4 + 2sqrt3; If I square the RHS I get (sqrt3 + 1)^2 = 3 + 2sqrt3 + 1 (exactly the same as (a + b)^2 = a^2 + 2ab + b^2) so it becomes 4 + 2sqrt3 = LHS i.e. since the squares of both sides are equal (both equal to 4 + 2sqrt3) then the two sides must be equal without squaring them, so sqrt(4 + 2sqrt3) = sqrt3 + 1. Hope I've made it clear enough.

Answer 5

What you need to do here is express 4 as (sqrt3)^2 + 1.



You can now rewrite the radicand as



(sqrt3)^2 + 2sqrt3 + 1, which equals



(sqrt3 + 1)^2. So



sqrt(4 + 2sqrt3) = sqrt(sqrt3 + 1)^2



= sqrt3 + 1