Fiendish! In the end I just factorised out sqrt(5) within the main radical leaving 6+2sqrt(5), easy to guess as the square of 1+sqrt(5). NB there is a radical missing at the end of Method 1, it does actually work.
*@ SyberMath* -- You can let a + b*sqrt(5) = 10 + 6sqrt(5). Square each side: Then a^2 + 5b^2 = 10 and 2ab = 6sqrt(5). ab = 3 or b = 3/a. From these we get a^4 - 10a^2 + 45 = 0. a^2 = 5 +/- 2isqrt(5) I'll use a^2 = 5 + 2isqrt(5). So, a = sqrt[5 + 2isqrt(5)]. b*sqrt(5) = 3sqrt(5)/a = 3sqrt(5)/sqrt[5 + 2isqrt(5)] = sqrt[5 - 2isqrt(5)]. a = (1/2 + i/2)5^(1/4)[sqrt(5) - i] and b = (1/2 + i/2)[5^(1/4) - i(125)^(1/4)]. a + b*sqrt(5) = 125^(1/4) + 5^(1/4)
This is known as _indirect denesting_ of a nested square root. Generally, a nested square root √(a ± b√c) where a, b, c are positive _rational_ numbers and √c is _irrational_ can be denested directly, that is, there exist positive rational numbers p and q such that √(a ± b√c) = √p ± √q, _if and only if_ a² − b²c is the square of a rational number. If √(a + b√c) and where a, b, c are rational numbers such that a + b√c is a positive real number and √c is irrational then we can take out a factor √c which gives √(a + b√c) = √(√c·(a/√c + b)) = √(√c)·√(a/√c + b) = ⁴√c·√(b + a/√c) We can now consider two cases. If a > 0 we have √(a + b√c) = ⁴√c·√(b + a/√c) = ⁴√c·√(b + √(a²/c)) and if a < 0 we have √(a + b√c) = ⁴√c·√(b + a/√c) = ⁴√c·√(b − √(a²/c)) Assuming b is positive, √(b ± √(a²/c)) can be denested directly, that is, there exist positive rational numbers p and q such that √(b ± √(a²/c)) = √p ± √q, _if and only if_ b² − a²/c is the square of a rational number and this is therefore the criterion for the _indirect denestibility_ of √(a + b√c) where a, b, c are rational numbers such that a + b√c is a positive real number and √c is irrational. For √(10 + 6√5) we have a = 10, b = 6, c = 5 and therefore a² − b²c = 100 − 36·5 = −80 which is not the square of a rational number. Therefore, √(10 + 6√5) cannot be denested directly, that is, there exist no positive rational numbers p and q such that √(10 + 6√5) = √p + √q. However, we _do_ have b² − a²/c = 36 − 100/5 = 16 which _is_ the square of a rational number. So, √(10 + 6√5) _can_ be denested indirectly, that is, there exist rational numbers p and q such that √(10 + 6√5) = ⁴√5·√(6 + 2√5) = ⁴√5·(√p + √q). Since 6² − (2√5)² = 36 − 20 = 16 is the square of a rational number, √(6 + 2√5) can be denested directly and is equal to √5 + 1, so we have √(10 + 6√5) = ⁴√5·√(6 + 2√5) = ⁴√5·(√5 + 1) = ⁴√5·(⁴√25 + 1) = ⁴√125 + ⁴√5 Note that since √(6 − 2√5) = √5 − 1 we also have √(−10 + 6√5) = ⁴√5·√(6 − 2√5) = ⁴√5·(√5 − 1) = ⁴√5·(⁴√25 − 1) = ⁴√125 − ⁴√5 For a full discussion, including proofs, I recommend reading the paper _On the denesting of nested square roots_ by Eleftherios Gkioulekas which is freely available and can easily be found if you google it. I cannot give a direct link here because RUclips will then remove this message.
Fiendish! In the end I just factorised out sqrt(5) within the main radical leaving 6+2sqrt(5), easy to guess as the square of 1+sqrt(5). NB there is a radical missing at the end of Method 1, it does actually work.
*@ SyberMath* -- You can let a + b*sqrt(5) = 10 + 6sqrt(5).
Square each side:
Then a^2 + 5b^2 = 10 and
2ab = 6sqrt(5).
ab = 3 or b = 3/a.
From these we get a^4 - 10a^2 + 45 = 0.
a^2 = 5 +/- 2isqrt(5)
I'll use a^2 = 5 + 2isqrt(5).
So, a = sqrt[5 + 2isqrt(5)].
b*sqrt(5) = 3sqrt(5)/a = 3sqrt(5)/sqrt[5 + 2isqrt(5)] = sqrt[5 - 2isqrt(5)].
a = (1/2 + i/2)5^(1/4)[sqrt(5) - i]
and b = (1/2 + i/2)[5^(1/4) - i(125)^(1/4)].
a + b*sqrt(5) = 125^(1/4) + 5^(1/4)
Wow! That’s good
This is known as _indirect denesting_ of a nested square root.
Generally, a nested square root √(a ± b√c) where a, b, c are positive _rational_ numbers and √c is _irrational_ can be denested directly, that is, there exist positive rational numbers p and q such that √(a ± b√c) = √p ± √q, _if and only if_ a² − b²c is the square of a rational number.
If √(a + b√c) and where a, b, c are rational numbers such that a + b√c is a positive real number and √c is irrational then we can take out a factor √c which gives
√(a + b√c) = √(√c·(a/√c + b)) = √(√c)·√(a/√c + b) = ⁴√c·√(b + a/√c)
We can now consider two cases. If a > 0 we have
√(a + b√c) = ⁴√c·√(b + a/√c) = ⁴√c·√(b + √(a²/c))
and if a < 0 we have
√(a + b√c) = ⁴√c·√(b + a/√c) = ⁴√c·√(b − √(a²/c))
Assuming b is positive, √(b ± √(a²/c)) can be denested directly, that is, there exist positive rational numbers p and q such that √(b ± √(a²/c)) = √p ± √q, _if and only if_ b² − a²/c is the square of a rational number and this is therefore the criterion for the _indirect denestibility_ of √(a + b√c) where a, b, c are rational numbers such that a + b√c is a positive real number and √c is irrational.
For √(10 + 6√5) we have a = 10, b = 6, c = 5 and therefore a² − b²c = 100 − 36·5 = −80 which is not the square of a rational number. Therefore, √(10 + 6√5) cannot be denested directly, that is, there exist no positive rational numbers p and q such that √(10 + 6√5) = √p + √q. However, we _do_ have b² − a²/c = 36 − 100/5 = 16 which _is_ the square of a rational number. So, √(10 + 6√5) _can_ be denested indirectly, that is, there exist rational numbers p and q such that √(10 + 6√5) = ⁴√5·√(6 + 2√5) = ⁴√5·(√p + √q).
Since 6² − (2√5)² = 36 − 20 = 16 is the square of a rational number, √(6 + 2√5) can be denested directly and is equal to √5 + 1, so we have
√(10 + 6√5) = ⁴√5·√(6 + 2√5) = ⁴√5·(√5 + 1) = ⁴√5·(⁴√25 + 1) = ⁴√125 + ⁴√5
Note that since √(6 − 2√5) = √5 − 1 we also have
√(−10 + 6√5) = ⁴√5·√(6 − 2√5) = ⁴√5·(√5 − 1) = ⁴√5·(⁴√25 − 1) = ⁴√125 − ⁴√5
For a full discussion, including proofs, I recommend reading the paper _On the denesting of nested square roots_ by Eleftherios Gkioulekas which is freely available and can easily be found if you google it. I cannot give a direct link here because RUclips will then remove this message.
I find it easier to remember that
_√(a ± b√n) = |t₁ ± t₂|_
where _t₁_ and _t₂_ are the roots of the quadratic
_t² - at + ¼b²n = 0_
Nice!
sqrt(6*5+2sqrt(125))/sqrt(5)=(5+sqrt(5))/sqrt(5)=sqrt(5)+1
=√((3+√5)^2-4)=√((5+√5)(1+√5))=√(√5(1+√5)(1+√5))=(√√5)(1+√5)
problem
De-nest √(10+6√5)
First let's simplify by factoring out√(2√5).
√(10+6√5) =√(2√5) √(3+√5)
After we de-nest √(3+√5) the answer will be the de-nested form of √(3+√5) times √(2√5).
De-nesting √(3+√5)
√(3+√5) = a + b √5
Square both sides.
3 + √5 = a² +5b² +2ab√5
3 = a² +5b²
2ab = 1
4 a² b² = 1
a² = 1/(4 b²)
3 = 1/(4 b²) + 5 b²
12 b² = 1 + 20 b⁴
20 b⁴ - 12 b² + 1 = 0
b ² = [ 12 ± √(144-4•20) ] / 40
= [ 12 ± 8 ] / 40
= 1/2, 1/10
a² = 1/(4 b²)
= 1/2, 5/2
a,b = 1/√2, 1/√2
√(3+√5) = 1/√2 + √5/√2
= (1+√ 5)/√2
This is a correct de-nesting of √(3+√5).
Therefore
√(10+6√5) =√(2√5) (1+√ 5)/√2
= ∜5 + ∜5 √5
is the de-nested form of √(10+6√5) .
answer
∜5 + ∜5 √5